Chapter 1

Sequences and Series

• Sequences and Series
• Convergence of Infinite Series
• Tests of Convergence
• P-Series Test
• Comparison Tests
• Ratio test
• Raabe’s test
• Cauchy’s Root test
• Integral test
• Leibnitz’s test
• Absolute Convergence
• Conditional convergence
• Power series and Interval of convergence
• Summary of all Tests
• Solved University Questions (JNTU)
• Objective type of Questions
2 Engineering Mathematics - I

1.1 Sequence
A function f:N → S, where S is any nonempty set is called a Sequence
i.e., for each n ∈ N, ∃ a unique element f(n) ∈ S. The sequence is written as f(1), f(2),
f(3), ......f(n)...., and is denoted by {f(n)}, or <f(n)>, or (f(n)). If f(n) = an , the sequence is
written as a1 , a2 .....an and denoted by , {an } or < an > or ( an ) . Here f(n) or an are the
nth terms of the Sequence.
Ex. 1. 1 , 4 , 9 , 16 ,......... n 2 ,.....(or) < n 2 >
N S
1
2 4
3 9
. .
. .
n n2
. .
. .

1 1 1 1 ⎛ 1 ⎞
Ex. 2. , 3 , 3 ,..... 3 ....(or ) ⎜ 3 ⎟
3
1 2 3 n ⎝n ⎠

Ex. 3. 1, 1, 1......1..... or <1>

Sequences and Series 3

( −1)
n −1
Ex 4: 1 , –1, 1, –1, ......... or

Note : 1. If S ⊆ R then the sequence is called a real sequence.

2. The range of a sequence is almost a countable set.

1.1.1 Kinds of Sequences

1. Finite Sequence: A sequence < an > in which an = 0 ∀n > m ∈ N is said to
be a finite Sequence. i.e., A finite Sequence has a finite number of terms.
2. Infinite Sequence: A sequence, which is not finite, is an infinite sequence.

1.1.2 Bounds of a Sequence and Bounded Sequence

1. If ∃ a number ‘M’ ∋ an ≤ M, ∀n ∈ N, the Sequence < an > is said to be
bounded above or bounded on the right.
1 1
Ex. 1, , , ,....... here an ≤ 1 ∀n ∈ N
2 3
2. If ∃ a number ‘m’ ∋ an ≥ m, ∀n ∈ N, the sequence < an > is said to be
bounded below or bounded on the left.
Ex. 1 , 2 , 3 ,.....here an ≥ 1 ∀n ∈ N
3. A sequence which is bounded above and below is said to be bounded.
n⎛ 1⎞
Ex. Let an = ( −1) ⎜ 1 + ⎟
⎝ n⎠

n 1 2 3 4 ......
an -2 3/2 -4/3 5/4 ......
4 Engineering Mathematics - I

3
From the above figure (see also table) it can be seen that m = –2 and M = .
2
∴ The sequence is bounded.
1.1.3 Limits of a Sequence
A Sequence < an > is said to tend to limit ‘l’ when, given any + ve number ' ∈ ',
however small, we can always find an integer ‘m’ such that an − l <∈, ∀n ≥ m , and we
write Lt an = l or an → l
n →∞

n2 + 1 1
Ex. If an = then < an >→ .
2n + 3
2
2

1.1.4 Convergent, Divergent and Oscillatory Sequences

1. Convergent Sequence: A sequence which tends to a finite limit, say ‘l’ is
called a Convergent Sequence. We say that the sequence converges to ‘l’
2. Divergent Sequence: A sequence which tends to ±∞ is said to be Divergent
(or is said to diverge).
3. Oscillatory Sequence: A sequence which neither converges nor diverges ,is
called an Oscillatory Sequence.
3 4 5 1
Ex. 1. Consider the sequence 2 , , , ,..... here an = 1 +
2 3 4 n
The sequence < an > is convergent and has the limit 1
1 1 1 1
an − 1 = 1 + − 1 = and <∈ whenever n >
n n n ∈
1
Suppose we choose ∈= .001 , we have < .001 when n > 1000.
n
1
Ex. 2. If an = 3 + ( −1)
n
< an > converges to 3.
'n'
Sequences and Series 5

If an = n 2 + ( −1) .n, < an > diverges.

n
Ex. 3.
1
+ 2 ( −1) , < an > oscillates between -2 and 2.
n
Ex. 4. If an =
n

1.2 Infinite Series

If < un > is a sequence, then the expression u1 + u2 + u3 + ........ + un + ..... is called an
∞
infinite series. It is denoted by Σ un or simply Σun
n =1

The sum of the first n terms of the series is denoted by sn

i.e., sn = u1 + u2 + u3 + ...... + un ; s1 , s2 , s3 ,....sn are called partial sums.

1.2.1 Convergent, Divergent and Oscillatory Series

Let Σun be an infinite series. As n → ∞, there are three possibilities.
(a) Convergent series: As n → ∞, sn → a finite limit, say ‘s’ in which case the
series is said to be convergent and ‘s’ is called its sum to infinity.
Thus Lt sn = s (or) simply Ltsn = s
n →∞
∞
This is also written as u1 + u2 + u3 + ..... + un + ...to∞ = s. (or) Σ un = s (or)
n =1

simply Σun = s.
(b) Divergent series: If sn → ∞ or −∞ , the series said to be divergent.
(c) Oscillatory Series: If sn does not tend to a unique limit either finite or infinite it
is said to be an Oscillatory Series.

Note: Divergent or Oscillatory series are sometimes called non convergent series.

1.2.2 Geometric Series

The series, 1 + x + x 2 + .....x n −1 + ... is
1
(i) Convergent when x < 1 , and its sum is
1− x
(ii) Divergent when x ≥ 1 .
(iii) Oscillates finitely when x = -1 and oscillates infinitely when x < -1.

Proof: The given series is a geometric series with common ratio ‘x’
1 − xn
∴ sn = when x ≠ 1 [By actual division – verify]
1− x
6 Engineering Mathematics - I

(i) When x < 1:

⎛ 1 ⎞ ⎛ xn ⎞ 1
Lt sn = Lt ⎜ ⎟ − Lt ⎜ ⎟ = ⎡⎣since x n → 0 as n → ∞ ⎤⎦
n →∞ 1 − x
n →∞
⎝ ⎠ n →∞ ⎝ 1 − x ⎠ 1 − x
1
∴ The series converges to
1− x
x −1
n
(ii) When x ≥ 1: sn = and sn → ∞ as n → ∞
x −1
∴ The series is divergent.
(iii) When x = –1: when n is even, sn → 0 and when n is odd, sn → 1
∴ The series oscillates finitely.
(iv) When x < −1, sn → ∞ or −∞ according as n is odd or even.
∴ The series oscillates infinitely.
1.2.3 Some Elementary Properties of Infinite Series
1. The convergence or divergence of an infinites series is unaltered by an addition or
deletion of a finite number of terms from it.
2. If some or all the terms of a convergent series of positive terms change their signs,
the series will still be convergent.
3. Let Σun converge to ‘s’
Let ‘k’ be a non – zero fixed number. Then Σkun converges to ks.
Also, if Σun diverges or oscillates, so does Σkun

4. Let Σun converge to ‘l’ and Σvn converge to ‘m’. Then

(i) Σ(un + vn ) converges to ( l + m ) and (ii) Σ(un + vn ) converges to ( l – m )

1.2.4 Series of Positive Terms

Consider the series in which all terms beginning from a particular term are +ve.
Let the first term from which all terms are +ve be u1 .
Let Σun be such a convergent series of +ve terms. Then, we observe that the
convergence is unaltered by any rearrangement of the terms of the series.

1.2.5 Theorem
If Σun is convergent, then Lt un = 0 .
n →∞

Proof : sn = u1 + u2 + ...... + un
sn −1 = u1 + u2 + ...... + un −1, , so that, un = sn − sn −1
Sequences and Series 7

Suppose Σun = l then Lt sn = l and Lt sn −1 = l

n →∞ n →∞

∴ Lt un = Lt ( sn − sn −1 ) ; Lt sn − Lt sn −1 = l − l = 0
n →∞ n →∞ n →∞ n →∞
Note: The converse of the above theorem need not be always true. This can be
observed from the following examples.
1 1 1 1
(i) Consider the series, 1 + + + ....... + + .... ; un = , Lt un = 0
2 3 n n n →∞
1
But from p-series test (1.3.1) it is clear that Σ is divergent.
n
1 1 1 1
(ii) Consider the series, + 2 + 2 + ..... + 2 + ......
2
1 2 3 n
1 1
un = , Lt u = 0, by p series test, clearly Σ 2 converges,
2 n →∞ n
n n
Note : If Lt un ≠ 0 the series is divergent;
n →∞

2n − 1
Ex. un = , here Lt un = 1 ∴ Σun is divergent.
2n n →∞

1.3 Tests for the Convergence of an Infinite Series

In order to study the nature of any given infinite series of +ve terms regarding
convergence or otherwise, a few tests are given below.

1.3.1 P-Series Test

1 ∞
1 1 1
The infinite series, Σ p
= p + p + p + ......, is
n 1n =12 3
(i) Convergent when p > 1, and (ii) Divergent when p ≤ 1 . (JNTU 2002, 2003)
Proof :
1 1
Case (i) Let p > 1; p > 1,3 p > 2 p ; ⇒ p
< p
3 2
1 1 1 1 2
∴ p
+ p < p+ p = p
2 3 2 2 2
1 1 1 1 1 1 1 1 4
Similarly, p
+ p + p+ p < p+ p + p+ p = p
4 5 6 7 4 4 4 4 4
1 1 1 8
p
+ p + .... + p < p , and so on.
8 9 16 8
Adding we get
8 Engineering Mathematics - I

1 2 4 8
Σ p
< 1 + p + p + p + ....
n 2 4 8
1 1 1 1
i.e., Σ p
< 1 + ( p −1) + 2( p −1) + 3( p −1) + ......
n 2 2 2
The RHS of the above inequality is an infinite geometric series with common
1
ratio p −1 < 1( since p > 1) The sum of this geometric series is finite.
2
∞ 1
Hence Σ is also finite.
n =1 n p

∴ The given series is convergent.

1 1 1 1
Case (ii) Let p =1; Σ = 1 + + + + ......
np 2 3 4
1 1 1 1 1
We have, + > + =
3 4 4 4 2
1 1 1 1 1 1 1 1 1
+ + + > + + + =
5 6 7 8 8 8 8 8 2
1 1 1 1 1 1 1
+ + ....... > + + ..... = and so on
9 10 16 16 16 16 2
1 ⎛1 1⎞ ⎛1 1 1 1⎞
∴ Σ p = 1 + ⎜ + ⎟ + ⎜ + + + ⎟ + .....
n ⎝ 2 3⎠ ⎝ 4 5 6 7 ⎠
1 1 1
≥ 1 + + + + .....
2 2 2
The sum of RHS series is ∞
⎛ n −1 n +1 ⎞
⎜ since sn = 1 + = and Lt sn = ∞ ⎟
⎝ 2 2 n →∞
⎠
∞ 1
∴ The sum of the given series is also ∞ ; ∴ Σ ( p = 1 ) diverges.
n =1 n p

1 1 1
Case (iii) Let p<1, Σp
= 1 + p + p + .....
n 2 3
1 1 1 1
Since p < 1, p > , p > , ...... and so on
2 2 3 3
1 1 1 1
∴ Σ p > 1 + + + + .......
n 2 3 4
From the Case (ii), it follows that the series on the RHS of above inequality is
divergent.
Sequences and Series 9

1
∴ Σ is divergent , when P < 1
np
Note: This theorem is often helpful in discussing the nature of a given infinite series.

1.3.2 Comparison Tests

1. Let Σun and Σvn be two series of +ve terms and let Σvn be convergent.
Then Σun converges,
(a) If un ≤ vn , ∀ n ∈ N
u
(b) or n ≤ k ∀n ∈ N where k is > 0 and finite.
vn
u
(c) or n → a finite limit > 0
vn
Proof : (a) Let Σvn = l (finite)
Then, u1 + u2 + ..... + un + ...... ≤ v1 + v2 + .....vn + ..... ≤ l > 0
Since l is finite it follows that Σun is convergent
un
(c) ≤ k ⇒ un ≤ kvn , ∀n ∈ N , since Σvn is convergent and k (>0) is finite,
vn
Σkvn is convergent ∴ Σun is convergent.
u u
(d) Since Lt n is finite, we can find a +ve constant k ,∋ n < k ∀n ∈ N
n →∞ v vn
n

∴ from (2) , it follows that Σun is convergent

2. Let Σun and Σvn be two series of +ve terms and let Σvn be divergent. Then
Σun diverges,
* 1. If un ≥ vn , ∀n ∈ N
u
or * 2. If n ≥ k , ∀n ∈ N where k is finite and ≠ 0
vn
u
or * 3. If Lt n is finite and non-zero.
n →∞ v
n
Proof:
1. Let M be a +ve integer however large it may be. Science Σvn is divergent, a
number m can be found such that
v1 + v2 + ..... + vn > M , ∀n > m
∴ u1 + u2 + ..... + un > M , ∀n > m ( un ≥ vn )
∴ Σun is divergent
10 Engineering Mathematics - I

2. u1 ≥ kvn ∀n
Σvn is divergent ⇒ Σkvn is divergent
∴ Σun is divergent
u u
3. Since Lt n is finite, a + ve constant k can be found such that n > k , ∀n
n →∞ v vn
n

(probably except for a finite number of terms )

∴ From (2), it follows that Σun is divergent.
Note :
(a) In (1) and (2), it is sufficient that the conditions with * hold ∀n > m ∈ N
Alternate form of comparison tests : The above two types of comparison tests
2.8.(1) and 2.8.(2) can be clubbed together and stated as follows :
un
If Σun and Σvn are two series of + ve terms such that Lt = k, where k is
n →∞ v
n
non- zero and finite, then Σun and Σvn both converge or both diverge.
(b) 1. The above form of comparison tests is mostly used in solving problems.
2. In order to apply the test in problems, we require a certain series Σvn whose
nature is already known i.e., we must know whether Σvn is convergent are
divergent. For this reason, we call Σvn as an ‘auxiliary series’.
3. In problems, the geometric series (1.2.2.) and the p-series (1.3.1) can be
conveniently used as ‘auxiliary series’.

Solved Examples
EXAMPLE 1
Test the convergence of the following series:
∞
(c) Σ ⎡( n 4 + 1) − n⎤
3 4 5 6 4 5 6 7 14
(a) + + + + ..... (b) + + + + .....
1 8 27 64 1 4 9 16 n =1 ⎢
⎣ ⎥⎦

SOLUTION
(a) Step 1: To find "un " the nth term of the given series. The numerators 3, 4, 5,
6......of the terms, are in AP.
nth term tn = 3 + ( n − 1) .1 = n + 2
n+2
Denominators are 13 , 23 ,33 , 43.....nth term = n3 ; ∴ un =
n3
Step 2: To choose the auxiliary series Σvn . In un , the highest degree of n in the
numerator is 1 and that of denominator is 3.
Sequences and Series 11

1 1
∴ we take, vn = 3−1
= 2
n n
un n+2 2 n+2 ⎛ 2⎞
Step 3: Lt = Lt × n = Lt = Lt ⎜1 + ⎟ = 1, which is non- zero and
n →∞ vn n →∞ n 3 n →∞ n n →∞
⎝ n⎠
finite.
un
Step 4: Conclusion: Lt =1
n →∞ v
n

1
∴ Σun and Σvn both converge or diverge (by comparison test). But Σvn = Σ is
n2
convergent by p-series test (p = 2 > 1); ∴ Σun is convergent.

4 5 6 7
(b) + + + + .....
1 4 9 16
n+3
Step 1: 4 , 5, 6, 7, .....in AP , tn = 4 + ( n − 1)1 = n + 3 ∴ un =
n2
1
Step 2: Let Σvn = be the auxiliary series
n
u ⎛ n +3⎞ ⎛ 3⎞
Step 3: Lt n = Lt ⎜ 2 ⎟ × n = Lt ⎜1 + ⎟ = 1 , which is non-zero and finite.
n →∞ v
n
n →∞
⎝ n ⎠ n →∞
⎝ n⎠
Step 4: ∴ By comparison test, both Σun and Σvn converge are diverge together.
1
But Σvn = Σ is divergent, by p-series test (p = 1); ∴ Σun is divergent.
n
1
⎡ 1 ⎤
∞ ⎧ ⎛ ⎞ ⎫ ⎛ ⎞
Σ ⎡( n + 1) ⎤
− n = ⎨n ⎜1 + 4 ⎟⎬ – n = n ⎜1 + 4 ⎟ – 1⎥
⎢
4 14
4 1 4 1 4
(c)
⎣⎢
n =1 ⎦⎥ ⎩ ⎝ n ⎠⎭ ⎢⎝ n ⎠ ⎥
⎣⎢ ⎦⎥
⎡ 1⎛1 ⎞ ⎤
⎢ ⎜ − 1⎟ ⎥ ⎡ 1 ⎤
= n ⎢1 + 4 + ⎝
1 4 4 ⎠ 1 3
. 8 + ..... − 1⎥ = n ⎢ 4 − + .....⎥
⎢ n n ⎥ ⎣ n n ⎦
8
4 2! 4 32
⎢⎣ ⎥⎦
1 3 1 ⎡1 3 ⎤
= 3− + .... = 3 ⎢ − + .....⎥
4n 32n 7 n ⎣ 4 32n 4 ⎦
1
Here it will be convenient if we take vn = 3
n
12 Engineering Mathematics - I

un ⎛1 1 ⎞ 1
Lt = Lt ⎜ − + ..... ⎟ = , which is non-zero and finite
n →∞ v 32n
⎝ ⎠ 4
n →∞ 4 4
n

∴ By comparison test, Σun and Σvn both converge or both diverge. But by p-
1
series test Σvn = is convergent. (p = 3 > 1); ∴ Σun is convergent.
n3

EXAMPLE 2
3
3n 2 + 1
If u n = show that Σun is divergent
4
2n3 + 3n + 5

SOLUTION
As n increases, un approximates to
1 2 1
3
3n 2 3 3
n 3
3 3
1
= 1
× 3
= 1
. 1
4
2n3 2 4
n 4
2 4
n 12

1
1 u 33
∴ If we take vn = , Lt n = 1 which is finite.
1 n →∞ v
n 12 n 2 4
1
[(or) Hint: Take vn = l1 − l2
, where l1 and l2 are indices of ‘n’ of the largest terms
n
1 1
in denominator and nominator respectively of un . Here vn = 3 2
= 1
]
−
n 4 3
n 12

1
By comparison test, Σvn and Σun converge or diverge together. But Σvn = Σ 1
is
n 12

1
divergent by p – series test ( since p = <1)
12
∴ Σun is divergent.

EXAMPLE 3
1 2 3 4
Test for the convergence of the series. + + + + ......
2 3 4 5
SOLUTION
n 1 1 u 1
Here, un = ; Take vn = = = 1 , Lt n = Lt = 1 (finite)
n +1 1 1
− n 0 n →∞ vn n →∞
1+
1
n 2 2
n
Sequences and Series 13

Σvn is divergent by p – series test. (p = 0 < 1)

∴ By comparison test, Σun is divergent, (Students are advised to follow the procedure
given in ex. 1.2.9(a) and (b) to find “ un ” of the given series.)

EXAMPLE 4
1 1 1
Show that 1 + + + ....... + + ..... is convergent.
1 2 n
SOLUTION
1
un = (neglecting 1st term )
n
1 1 1
= < = n −1
1.2.3......n 1.2.2.2.....n − 1times (2 )
1 1 1
∴ Σun < 1 + + 2 + 3 + ......
2 2 2
1
which is an infinite geometric series with common ratio < 1
2
1
∴ Σ n −1 is convergent. (1.2.3(a)). Hence Σun is convergent.
2

EXAMPLE 5
1 1 1
Test for the convergence of the series, + + + .......
1.2.3 2.3.4 3.4.5
SOLUTION
1 1 un
n3
un = ; Take vn = Lt = Lt = 1 (finite)
n ( n + 1)( n + 2 ) n3 3⎛
n →∞ v1 ⎞⎛ 2 ⎞
n →∞
n ⎜ 1 + ⎟⎜ 1 + ⎟
n

⎝ n ⎠⎝ n ⎠
∴ By comparison test, Σun , and Σvn converge or diverge together. But by p-series test,
1
Σvn = Σ 3 is convergent ( p = 3 > 1 ); ∴ Σun is convergent .
n
EXAMPLE 6

If un = n 4 + 1 − n 4 − 1, show that Σun is convergent. [JNTU, 2005]

SOLUTION
1 1
⎛ 1 ⎞2 ⎛ 1 ⎞2
un = n 2 ⎜ 1 + 4 ⎟ − n 2 ⎜ 1 − 4 ⎟
⎝ n ⎠ ⎝ n ⎠
14 Engineering Mathematics - I

⎡⎛ 1 1 1 ⎞ ⎛ 1 1 1 ⎞⎤
= n 2 ⎢⎜ 1 + 4 − 8 + − .... ⎟ − ⎜1 − 4 − 8 − − ..... ⎟ ⎥
⎣⎝ 2n 8n 16n ⎠ ⎝ 2n 8n 16n ⎠⎦
12 12

⎡1 1 ⎤ 1 ⎡ 1 ⎤
= n 2 ⎢ 4 + 12 + ....⎥ = 2 ⎢1 + 10 + ....⎥
⎣ n 8n ⎦ n ⎣ 8n ⎦
1 u
Take vn = 2 , hence Lt n = 1
n n →∞ vn
1
∴ By comparison test, Σun and Σvn converge or diverge together. But Σvn = is
n2
convergent by p –series test (p = 2 > 1) ∴ Σun is convergent.

EXAMPLE 7
1 1 1
Test the series + + + ..... for convergence.
1+ x 2 + x 3 + x
SOLUTION
1 1 un n 1
un = ; take vn = , then = =
n+ x n vn n + x 1 + x
n
⎛ ⎞
⎜ 1 ⎟ 1
Lt ⎜ ⎟ = 1; Σvn = Σ is divergent by p-series test (p =1 )
⎜1+ x ⎟ n
n →∞

⎝ n⎠
∴ By comparison test, Σun is divergent.
EXAMPLE 8
∞
⎛1⎞
Show that Σ sin ⎜ ⎟ is divergent.
n =1
⎝n⎠
SOLUTION
⎛1⎞ 1
un = sin ⎜ ⎟ ; take vn =
⎝n⎠ n
⎛1⎞
sin ⎜ ⎟
u ⎝ n ⎠ = Lt sin t (where t = 1 ) = 1
Lt n = Lt
n →∞ v n →∞ ⎛ 1 ⎞ t →0 t n
n
⎜ ⎟
⎝n⎠
1
∴ Σu , Σvn both converge or diverge . But Σvn = Σ is divergent
n
n
( p -series test, p = 1 ); ∴ Σun is divergent.
Sequences and Series 15

EXAMPLE 9
⎛1⎞
Test the series Σ sin −1 ⎜ ⎟ for convergence.
⎝n⎠
SOLUTION
1 1
un = sin −1 ; Take vn =
n n
⎛1⎞
−1⎜ ⎟
u sin ⎝ n ⎠ ; = ⎛ θ ⎞ ⎛ 1 ⎞
Lt n = Lt Lt ⎜ ⎟ = 1⎜ Taking sin −1 = θ ⎟
n →∞ v
n
n →∞ ⎛ 1 ⎞ θ → 0
⎝ sin θ ⎠ ⎝ n ⎠
⎜ ⎟
n
⎝ ⎠
But Σvn is divergent. Hence Σun is divergent.

EXAMPLE 10
1 22 33
Show that the series 1 + + + + ..... is divergent.
22 33 43
SOLUTION
1 2 2 33
Neglecting the first term, the series is + + + ..... . Therefore
2 2 33 4 4
nn nn nn 1
un = = n= = ;
( n + 1)
n +1
( )( )
n + 1 n + 1 ⎛ 1⎞ n⎛ 1⎞
n
⎛ 1 ⎞⎛ 1 ⎞
n

n ⎜1 + ⎟ .n ⎜1 + ⎟ n ⎜ 1 + ⎟⎜1 + ⎟
⎝ n⎠ ⎝ n⎠ ⎝ n ⎠⎝ n ⎠
1
Take vn =
n
un 1 1 1
∴ Lt = Lt n
= Lt n
=
vn
n →∞ n →∞
⎛ 1 ⎞⎛ 1 ⎞ n →∞
⎛ 1⎞ e
⎜ 1 + ⎟⎜1 + ⎟ ⎜1 + ⎟ .1
⎝ n ⎠⎝ n ⎠ ⎝ n⎠
1
which is finite and Σvn = Σ is divergent by p –series test ( p = 1)
n
∴ Σun is divergent.

EXAMPLE 11
1 3 5
Show that the series + + + .......∞ is convergent. (JNTU 2000)
1.2.3 2.3.4 3.4.5
SOLUTION
1 3 5
+ + + .......∞
1.2.3 2.3.4 3.4.5
16 Engineering Mathematics - I

⎛ 1⎞
⎜ 2− ⎟
2n − 1 1 ⎝ n⎠
nth term = un = = 2.
n ( n + 1)( n + 2 ) n ⎛ 1 ⎞ ⎛ 2 ⎞
⎜1 + ⎟ ⎜1 + ⎟
⎝ n ⎠⎝ n ⎠
1
Take vn = 2
n
⎛ 1⎞
⎜ 2− ⎟
u 1 ⎝ n⎠ ⎛ 1 ⎞
Lt n = Lt 2 ÷⎜ 2 ⎟
n →∞ v
n
n →∞ n
1+ 1 1+ 2
n ( n )(
⎝n ⎠ )
u 2−0
Lt n = = 2 which is finite and non-zero
n →∞ v
n (1 + 0 )(1 + 0 )
∴ By comparison test ∑u n ∑ v converge or diverge together
and n

∑v is convergent. ∴ ∑ u is also convergent.

1
But n = ∑n 2 n

EXAMPLE 12
∞
1
Test whether the series ∑
n =1 n + n +1
is convergent (JNTU 1997, 1999, 2003)

SOLUTION
∞
1
The given series is ∑
n + n +1
n =1

1
un =
n + n +1
n +1 − n
= n +1 − n
( )( )
=
n + n +1 n +1 − n

⎪⎧⎛ 1 ⎞ 2 ⎪⎫
1
⎧⎛ 1 1 ⎞ ⎫
un = n ⎨⎜1 + ⎟ − 1⎬ = n ⎨⎜1 + − 2 + ..... ⎟ − 1⎬
n⎠ ⎩⎝ 2n 8n
⎩⎪⎝ ⎭⎪ ⎠ ⎭
⎧1 1 ⎫ 1 ⎧1 1 ⎫
u n = n ⎨ – 2 + ...⎬ = ⎨ – + ....⎬
⎩ 2 n 8n ⎭ n ⎩ 2 8n ⎭
Sequences and Series 17

1
Take vn =
n
un 1 ⎧1 2 ⎫ ⎛ 1 ⎞ 1
Lt = Lt ⎨ − + ......⎬ ÷ ⎜ ⎟=
n →∞ v n ⎩ 2 8n
n
n →∞
⎭ ⎝ n⎠ 2
which is finite and non-zero .
Using comparison test ∑ un and ∑v n converge or diverge together.

But
1
∑v = ∑ n
n (
is divergent since p = 1
2 )
∴ ∑u n is also divergent.

EXAMPLE 13
∞
Test for convergence ∑ ⎡⎣
n =1
3
n3 + 1 − n ⎤
⎦
[JNTU 1996, 2003, 2003]

th
n term
⎡⎛ 1⎞3 ⎤
1 ⎡
⎢ 1
un = n ⎢⎜ 1 + 3 ⎟ − 1⎥ = n 1 + 3 +
1 1 −1
3 3 1
⎤
. 6 + ..... − 1⎥
( )
⎢⎣⎝ n ⎠ ⎥⎦ ⎢ 3n 1.2 n ⎥
⎣ ⎦
1 1 1 ⎛1 1 ⎞ 1
= − 5 + ...... = 2 ⎜ − 3 + ...... ⎟ ; Let vn = 2
3n 9n
2
n ⎝ 3 9n ⎠ n
Then Lt
n →∞
un
vn n →∞ 3 9n (
= Lt 1 − 1 3 + .... = 1 ≠ 0
3 )
∴ By comparison test, ∑u n and ∑v n both converge or diverge.
But ∑v n is convergent by p -series test ( since p = 2 > 1) ∴ ∑u n is convergent.

EXAMPLE 14
2 3 4
Show that the series, p
+ p + p + ....... is convergent for p > 2 and divergent for
1 2 3
p≤2
SOLUTION

nth term of the given series = un =

n +1 n 1+ n
=
1 (
=
1+ 1) (n )
p p p −1
n n n
1 un
Let us take vn = ; Lt = 1 ≠ 0;
n p −1 n →∞ vn
18 Engineering Mathematics - I

∴ ∑ u and ∑ v
n n both converge or diverge by comparison test.
But ∑ v = ∑ 1
n p −1 converges when p -1>1 ; i.e., p >2 and diverges when
n
p − 1 ≤ 1 i.e p ≤ 2 ; Hence the result.
EXAMPLE 15
1
⎛ 2n + 3 ⎞
∞ 2
Test for convergence ∑ ⎜ n ⎟ (JNTU 2003)
n =1 ⎝ 3 + 1 ⎠

SOLUTION

( )
1
⎡ 2n 1 + 3 ⎤ 2
1
⎛ ⎞
un ⎜ 1 +
2
n 3
⎢ 2n ⎥ 2 n
2 ⎟
un = ⎢ vn = n ; =
( ) ⎥ ; Take
vn ⎜ 1 + 1 n⎟
⎢⎣ 3 1 + 3n
n 1 3
⎥⎦ ⎝ 3 ⎠
un
Lt
n →∞ v
= 1 ≠ 0 ; ∴ By comparison test, ∑u n and ∑v n behave the same way.
n
n 3
∞
⎛2⎞ 2 2 2 ⎛2⎞ 2
But ∑ vn = ∑ ⎜ ⎟ = + + ⎜ ⎟ + ....., which is a geometric series with
n =1 ⎝ 3 ⎠ 3 3 ⎝3⎠
common ratio 2 (<1) ∴ ∑ vn is convergent. Hence ∑ un is convergent.
3
EXAMPLE 16
1 4 9
Test for convergence of the series, + + + ...... (JNTU 2003)
4.7.10 7.10.13 10.13.16
SOLUTION
4, 7, 10,..............is an A . P; tn = 4 + ( n − 1) 3 = 3n + 1
7, 10, 13,............is an A . P; tn = 7 + ( n − 1) 3 = 3n + 4
and 10 , 13 , 16 ,.............is an A. P; tn = 10 + ( n − 1) 3 = 3n + 7
n2 n2
∴ un = =
( 3n + 1)( 3n + 4 )( 3n + 7 ) 3n 1 + 1 ( 3n ) .3n (1 + 4 3n ) .3n (1 + 7 3n )
1
( )( )( )
= ;
27 n 1 + 1 1+ 4 1+ 7
3n 3n 3n
Sequences and Series 19

1
Taking vn = , we get
n
un 1
Lt
n →∞ v
=
27
≠ 0 ; ∴ By comparison test, both ∑u n and ∑v n behave in the same
n

manner. But by p –series test, ∑v n is divergent, since p = 1. ∴ ∑u n is divergent .

EXAMPLE 17
2 n 2 − 5n + 1
Test for convergence ∑ 4n 3 − 7 n 2 + 2 (JNTU 2003)

SOLUTION
2 n 2 − 5n + 1
nth term of the given series = un =
4n3 − 7 n 2 + 2
1
Let vn =
n2
⎡ n 2− 5 + 1 ⎤ ⎡ 2− 5 + 1 ⎤
un ⎢ n n 2 n2 ⎥ ⎢ n n2 ⎥ 2
Lt = Lt . ⎢ × ⎥ = Lt ⎢ ⎥= 4 ≠0
n →∞ v
n
n →∞
⎢⎣ n 4 − n + n3
3 7 2
( 1
⎥⎦
n →∞ 7
)
⎢⎣ 4 − n + n3
2
( ) ⎥⎦
∴ By comparison test, ∑ un and ∑ vn both converge or diverge.
But ∑v n is convergent. [p series test – p = 2 > 1] ∴ ∑u n is convergent .

EXAMPLE 18
1
Test the series ∑u n , whose nth term is
( 4n 2
− i)
SOLUTION
⎡ ⎤
1 un ⎢
1 n2 ⎥ 1
un = Lt Let vn = 2 ,
= Lt ⎢ ⎥= 4≠0
( )
;
( 4n − i )
2 n →∞ v
n
n →∞ n i
⎢⎣ n 4 − n 2 ⎥⎦
2

∴ ∑ un and ∑ vn both converge or diverge by comparison test. But ∑ vn is

convergent by p –series test ( p = 2 > 1) ; ∴ ∑u n is convergent.
∞

∑ 4n
1
Note: Test the series
n =1
2
–1
20 Engineering Mathematics - I

EXAMPLE 19
⎛1⎞ ⎛1⎞
If un = ⎜ ⎟ .sin ⎜ ⎟ , show that
⎝n⎠ ⎝n⎠
∑u n is convergent .

SOLUTION
1
Let vn =
n2
, so that ∑v n is convergent by p –series test .

⎛u
Lt ⎜ n
⎞
= Lt
sin 1 ( )
n = Lt ⎛ sin t ⎞
⎟ n →∞ ⎜ ⎟
n →∞ v
⎝ n ⎠ 1
n ( )
t →0
⎝ t ⎠

⎛ un ⎞
where t = 1/n, Thus Lt ⎜ ⎟ =1≠ 0
n →∞
⎝ vn ⎠
∴ By comparison test, ∑u n is convergent.

EXAMPLE 20
1
Test for convergence ∑ n
tan( 1 )
n
SOLUTION
; Lt ⎡ n ⎤ = 1 ≠ 0 ( as in above example)
u
Take vn = 1
n
3
2 n →∞ ⎢⎣ vn ⎥⎦
Hence by comparison test, ∑u n converges as ∑v n converges.

EXAMPLE 21
∞
⎛1⎞
Show that ∑ sin
n =1
2
⎜ ⎟ is convergent.
⎝n⎠
SOLUTION

( ) ⎤⎥
2
⎡ sin 1
⎛ un ⎞
2
⎛1⎞ 1 n ⎛ sin t ⎞
Let un = sin ⎜ ⎟ ; 2
Take vn = 2 , Lt ⎜ ⎟ = Lt ⎢ = Lt ⎜ ⎟
n →∞ ⎢ ⎥
⎝n⎠ n n →∞ v
⎝ n⎠ 1 t → 0
⎝ t ⎠
⎣ n ⎦
t = 1 ; Lt ⎛⎜ n ⎞⎟ = 12 = 1 ≠ 0
u
where
n n→∞ ⎝ vn ⎠
∴ By comparison test, ∑ un and ∑ vn behave the same way.
But ∑v n is convergent by p- series test, since p = 2 > 1; ∴ ∑u n is convergent.
Sequences and Series 21

EXAMPLE 22
∞
Show that ∑
n=2
1
log ( n n )
is divergent.

SOLUTION
un = 1 ; log 2 < 1 ⇒ 2 log 2 < 2 ⇒ 1 >1 ;
n log n 2 log 2 2
Similarly 1 > 1 ..... 1 > 1 ,n∈ N
3log 3 3, n log n n
∴ ∑ 1 > ∑ 1 ; But ∑ 1 is divergent by p-series test.
n log n n n
By comparison test, given series is divergent. [If ∑ vn is divergent and un ≥ vn ∀n then

∑u n is divergent.]
(Note : This problem can also be done using Cauchy’s integral Test.
EXAMPLE 23
∞

∑ (c + n) ( d + n)
−r −s
Test the convergence of the series , where c, d, r, s are all +ve.
n =1

SOLUTION
1
The nth term of the series = un = .
(c + n) ( d + n)
r s

1 un nr +s 1
Let vn = r +s
Then = r s
= r s
n vn ⎛ c⎞ ⎛ d⎞ ⎛ c⎞ ⎛ d⎞
n ⎜ 1 + ⎟ .n s ⎜1 + ⎟
r
⎜1 + ⎟ ⎜1 + ⎟
⎝ n⎠ ⎝ n⎠ ⎝ n⎠ ⎝ n⎠
un
Lt = 1 ≠ 0 , ∴ ∑ un and ∑v n both converge are diverge, by comparison test.
n →∞ vn
But by p-series test, ∑v n converges if (r + s) > 1 and diverges if (r + s) ≤ 1
∴ ∑u n converges if ( r + s ) > 1 and diverges if ( r + s ) ≤ 1.

EXAMPLE 24
∞
( ) is divergent.
∑n
− 1+ 1
n
Show that
1

SOLUTION
un = n
(
− 1+ 1
n )= 1 Take vn =
1 u 1
; Lt n = Lt 1 = 1 ≠ 0
1
n.n n n n →∞ vn n →∞
n n
22 Engineering Mathematics - I

1
1 1
For let Lt = y say; log y = Lt − .log n = − Lt n = 0
n →∞ 1 n →∞ n n →∞ 1
n n

⎛∞⎞
∴ y = e0 = 1 (⎜
⎟ using L Hospitals rule)
⎝∞⎠
By comparison test both ∑ un and ∑ vn converge or diverge. But p-series test,

∑v n diverges (since p =1); Hence ∑u n diverges.

EXAMPLE 25
(n + a)
r
∞
Test for convergence the series ∑
n =1 ( n + b ) ( n + c )
p q
, a, b, c , p, q, r, being +ve.

SOLUTION

( ) ( )
r
1+ a
r
(n + a ) r nr 1 + a 1 n
un = = n
(1 + b n ) n (1 + c n )
= . ;
(n + b ) p (n + c )q ( )( )
p q p+ q −r p q
np q n 1+ b 1+ c
n n
1 un
Take vn = p+ q −r
; Lt =1≠ 0 ;
n n →∞ v
n

Applying comparison tests both ∑u n and ∑v n converge or diverge.

But by p-series test, ∑v n converges if ( p+ q –r ) > 1and diverges if ( p + q –r) ≤ 1.
Hence ∑u n converges if ( p + q – r ) > 1 and diverges if ( p + q – r) ≤ 1.

EXAMPLE 26
Test the convergence of the following series whose nth terms are:
( 3n + 4 ) 1 ⎛ 1 ⎞ ⎛ n +1 ⎞
n

(a) ; (b) tan ; (c) ⎜ 2 ⎟⎜ ⎟

( 2n + 1)( 2n + 3)( 2n + 5) n ⎝ n ⎠⎝ n + 3 ⎠
1 1
(d) ; (e)
( 3 + 5n )
n
n.3n
SOLUTION
1 ⎛u ⎞ 3
2 ∑ n
(a) Hint : Take vn = ; v is convergent; Lt ⎜ n ⎟ = ≠ 0 (Verify)
n n →∞ v
⎝ n⎠ 8
Apply comparison test:
∑ un is convergent [the student is advised to work out this problem fully ]
Sequences and Series 23

(b) Proceed as in Example 8; ∑u n is convergent.

(1 + 1n ) = e = 1 ≠ 0
n
1 ⎛u ⎞
(c) Hint : Take vn = 2 ; Lt ⎜ n ⎟ = Lt
( )
n
n n →∞ v
⎝ n ⎠ n→∞ 1 + 3 e3 e 2
n
1
vn = 2 is convergent (work out completely for yourself )
n
1 1 1 1 ⎛u ⎞
(d) un = = n. ; Take vn = n ; Lt ⎜ n ⎟ = 1 ≠ 0
3 +5
n n
5 ⎡ ⎛3⎞ ⎤ n
5 n →∞ v
⎝ n⎠
⎢1 + ⎜ ⎟ ⎥
⎣⎢ ⎝ 5 ⎠ ⎦⎥
∑u n and ∑v n behave the same way. But ∑v n is convergent since it is a
1
geometric series with common ratio <1
5
∴ ∑u n is convergent by comparison test .

1 1
(e) n
≤ n , ∀n ∈ N , since n.3n ≥ 3n ;
n.3 3
1 1
∴ ∑ n.3n ≤ ∑ 3n …..(1)

The series on the R.H .S of (1) is convergent since it is geometric series with
1
r= <1.
3
1
∴ By comparison test ∑ n.3 n
is convergent.

EXAMPLE 27
Test the convergence of the following series.

1+ 2 1+ 2 + 3 1+ 2 + 3 + 4
(a) 1+ + 2 + 2 + ...............
1 + 2 1 + 2 + 3 1 + 22 + 32 + 42
2 2 2 2

12 + 22 12 + 22 + 32 12 + 22 + 32 + 42
(b) 1 + 3 3 + 3 3 3 + 3 3 3 3 + ..............
1 +2 1 +2 +3 1 +2 +3 +4
24 Engineering Mathematics - I

SOLUTION

n
( n + 1)
1 + 2 + 3 + .... + n 2 3
(a) un = 2 = =
1 + 22 + 32 + .....n 2
n ( n + 1)
( 2n + 1) ( 2n + 1)
6
1 un ⎛ 3 n ⎞ 3
Take vn = ; Lt = Lt ⎜ ⎟ = ≠0
n n →∞ v n →∞ 2n + 1
n ⎝ ⎠ 2
∑u n and ∑v n behave alike by comparison test.
But ∑v n is diverges by p-series test. Hence ∑u n is divergent.

n ( n + 1)
( 2n + 1)
1 + 2 + .... + n
2 2 2
6 2 ( 2n + 1)
(b) un = = =
13 + 23 + .....n3 ( n + 1)
2
3n ( n + 1)
n2
4
1
Hint : Take vn = and proceed as in (a) and show that
n
∑u n is divergent.

Exercise 1.1
1. Test for convergence the infinite series whose nth term is:
1
(a) [Ans : divergent]
n− n
n +1 − n
(b) [Ans : convergent]
n
(c) n2 + 1 − n [Ans : divergent]
n
(d) [Ans : convergent]
n −1
2

(e) n3 + 1 − n3 [Ans : divergent]

1
(f) [Ans : divergent]
n ( n + 1)
n
(g) [Ans : convergent]
n +1
2

2n 3 + 5
(h) [Ans : convergent]
4n 5 + 1
Sequences and Series 25

2. Determine whether the following series are convergent or divergent.

1 2 3
(a) −1
+ −2
+ + ............ [Ans : divergent]
1+ 3 1+ 3 1 + 3−3
12 22 32 2 + 10n
(b) + 3 + 3 + ........ + + ..... ....... [Ans : convergent]
13
2 3 n3
1 1 1
(c) + + + ...... ........ [Ans : divergent]
1+ 2 2+ 3 3+ 4
2 3 4
(d) + + + ...... ......... [Ans : divergent]
32 42 52
1 1 1
(e) + + + .......... [Ans : convergent]
12 23 34
∞ 3
n2 + 1
(f) ∑ ............... [Ans : divergent]
n =1
4
4n 2 + 2 n + 3

∑ (8 )
∞ 1
(g) n
− 1 .................... [Ans : divergent]
1

∞
3n3 + 8
(h) ∑1 5n5 + 9 ........................ [Ans : convergent]

1 2 3
(i) + + + ........... [Ans : divergent]
1.3 3.5 5.7

1.3.3 D’ Alembert’s Ratio Test

un +1
Let (i) ∑u n be a series of +ve terms and (ii) Lt
n →∞ un
= k ( ≥ 0)

Then the series ∑u n is (i) convergent if k < 1 and (ii) divergent if k > 1.
Proof :
un +1
Case (i) Lt = k ( < 1)
n →∞ u
n
From the definition of a limit, it follows that
un +1
∃m > 0 and l ( 0 < l < 1) ∋ < l∀n ≥ m
un
26 Engineering Mathematics - I

um +1 u
i.e., < l , m + 2 < l ,..........
um um +1
⎡ u u ⎤
∴ um + um +1 + um + 2 + ...... ........ = um ⎢1 + m +1 + m + 2 + .....⎥
⎣ um um ⎦
um 1 um um 1
um 1 2
. .....
um um 1 um

< um (1 + l + l 2 + ...) = um .
1
( l < 1)
1− l
∞
1
But um .
1− l
is a finite quantity ∴ ∑u
n=m
n is convergent

By adding a finite number of terms u1 + u2 + ...... + um −1 , the convergence of the

∞
series is unaltered. ∑u
n=m
n is convergent.

un +1
Case (ii) Lt = k >1
n →∞ u
n

There may be some finite number of terms in the beginning which do not satisfy
un +1
the condition ≥ 1 . In such a case we can find a number ‘m’
un
un +1
∋ ≥ 1, ∀n ≥ m
un
Omitting the first ‘m’ terms, if we write the series as u1 + u2 + u3 + ........., we

have
u2 u u
≥ 1, 3 ≥ 1, 4 ≥ 1 .......... and so on
u1 u2 u3
⎛ u u u ⎞
∴ u1 + u2 + ...... + un = u1 ⎜1 + 2 + 3 . 2 + ..... ⎟ (to n terms)
⎝ u1 u2 u1 ⎠
≥ u1 (1 + 1 + 1.1 +.......to n terms)
= nu1
n
Lt
n →∞
∑u
n =1
n ≥ Lt n.u1 which → ∞ ; ∴
n →∞
∑u n is divergent .
Sequences and Series 27

∞
1
Note: 1 The ratio test fails when k = 1. As an example, consider the series, ∑n
n =1
p

p
p ⎛ 1 ⎞
un +1 ⎛ n ⎞ ⎜ ⎟ =1
Here Lt = Lt ⎜ ⎟ = Lt
n →∞ n + 1 n →∞ ⎜
n →∞ u
⎝ ⎠ 1+ 1 ⎟
n
⎝ n⎠
i.e., k = 1 for all values of p,
But the series is convergent if p > 1 and divergent if p ≤ 1 , which shows that
when k = 1, the series may converge or diverge and hence the test fails .
Note: 2 Ratio test can also be stated as follows:
un
If ∑u n is series of +ve terms and if Lt
n →∞ u
= k , then ∑u n is convergent
n +1

If k > 1 and divergent if k < 1 (the test fails when k = 1).

Solved Examples
Test for convergence of Series

EXAMPLE 28
x x2 x3
(a) + + + ................. (JNTU 2003)
1.2 2.3 3.4

SOLUTION
xn x n +1 un +1 x n +1 n ( n + 1) 1
un = ; un +1 = ; = . = x.
n ( n + 1) ( n + 1)( n + 2 ) un ( n + 1)( n + 2 ) x n
⎛ 2⎞
⎜1 + ⎟
⎝ n⎠
un +1
Therefore Lt =x
n →∞ u
n

∴ By ratio test ∑u n is convergent When |x| < 1 and divergent when | x | > 1;
1 1 u
When x = 1, un = ; Take vn = 2 ; Lt n = 1
n (1 + 1 n )
2
n n→∞ vn
∴ By comparison test ∑u n is convergent.
Hence ∑u n is convergent when x ≤ 1 and divergent when x > 1 .
28 Engineering Mathematics - I

(b) 1 + 3x + 5 x 2 + 7 x 3 + .......

SOLUTION
un +1 ⎛ 2n + 1 ⎞
un = ( 2n − 1) x n −1 ; un +1 = ( 2n + 1) x n ; Lt = Lt ⎜ ⎟x = x
n →∞ u n →∞ 2n − 1
n ⎝ ⎠
∴ By ratio test ∑u n is convergent when x < 1 and divergent when x > 1
When x = 1: un = 2n − 1; Lt un = ∞ ; ∴
n →∞
∑u n is divergent.

Hence ∑u n is convergent when x < 1 and divergent when x ≥ 1

∞
xn
(c) ∑
n =1 n + 1
2
...........

SOLUTION
xn x n +1
un = ; un +1 = .
n2 + 1 ( n + 1) +1
2

( )
⎡ 2 ⎤
un +1 ⎛ n + 1 ⎞ 2
un +1 ⎢ n 1 + 1 2 ⎥
=⎜ 2 = Lt ⎢ n ⎥ ( x) = x
Hence ⎟ x , nLt
un ⎝ n + 2n + 2 ⎠ →∞ u
n
n →∞
⎢ n2 1 + + 2 ⎞ ⎥
⎛ 2
⎢⎣ ⎜⎝ n n 2 ⎟⎠ ⎥⎦
∴ By ratio test, ∑ un is convergent when x < 1 and divergent when x > 1 When
1 1
x = 1: un = ; Take vn = 2
n +1 2
n
∴ By comparison test, ∑ un is convergent when x ≤ 1 and divergent when x > 1

EXAMPLE 29
∞
⎛ n2 − 1 ⎞ n
Test the series ∑ ⎜ 2 ⎟x , x > 0 for convergence.
n →∞ ⎝ n + 1 ⎠

SOLUTION
⎛ n2 − 1 ⎞ n ⎡ ( n + 1)2 − 1 ⎤ n +1
un = ⎜ 2 ⎟ x ; un +1 = ⎢ ⎥x
⎝ n +1 ⎠ ⎢⎣ ( n + 1) + 1 ⎥⎦
2
Sequences and Series 29

un +1 ⎡⎛ n 2 + 2n ⎞ ( n 2 + 1) ⎤
Lt = Lt ⎢⎜ 2 ⎟ 2 ⎥ .x
n →∞ u
n
n →∞
⎢⎣⎝ n + 2n + 2 ⎠ ( n − 1) ⎥⎦
⎡
= Lt ⎢ 4
(
n 4 (1 + 2 n ) 1 + 1 n 2 ⎤
⎥=x
)
⎣ (
n →∞ ⎢ n 1 + 2 n + 2 n 2 1 –)(
1 n 2
⎥⎦ )
∴ By ratio test, ∑ un is convergent when x < 1 and divergent when x > 1 when x = 1,
n2 − 1 1
un = Take vn = 0
n +1
2
n
Applying p-series and comparison test, it can be seen that ∑u n is divergent when x = 1.
∴ ∑u n is convergent when x < 1 and divergent x ≥ 1

EXAMPLE 30
2 p 3p 4 p
Show that the series 1 + + + + ..... , is convergent for all values of p.
2 3 4
SOLUTION
( n + 1)
p
np
un = ; un +1 =
n n +1
un +1 ⎡ ( n + 1) p n ⎤ ⎧⎪ 1 ⎛ n + 1 ⎞ p ⎫⎪
Lt = Lt ⎢ × p ⎥ = Lt ⎨ ⎜ ⎟ ⎬
n →∞ u
n
n →∞
⎢⎣ n + 1 n ⎥⎦ n→∞ ⎩⎪ ( n + 1) ⎝ n ⎠ ⎭⎪
p
1 ⎛ 1⎞
= Lt × Lt ⎜1 + ⎟ = 0 < 1 ;
n →∞ ( n + 1) n →∞
⎝ n⎠
∑u n is convergent for all ‘ p ‘ .

EXAMPLE 31
Test the convergence of the following series
1 1 1 1
p
+ p + p + p + ............
1 3 5 7
SOLUTION
1 1
un = ; un +1 =
( 2n − 1) ( 2n + 1)
p p
30 Engineering Mathematics - I

un +1 ( 2n − 1) 2 p.n p (1 − 1 2n )
p p
un +1
= = ; Lt =1
( 2n + 1) 2 p n p (1 + 1 2n )
p p
un n →∞ u
n

∴ Ratio test fails.

1 u np 1 un 1
Take vn = p ; n = = ; Lt = p,
n vn ( 2n − 1) p ⎛ 1 ⎞
p n →∞ v 2
2 p ⎜1 − ⎟ n

⎝ 2n ⎠
which is non – zero and finite
∴ By comparison test, ∑ un and ∑v n both converge or both diverge.
1
But by p – series test, ∑ v n = ∑n p
converges when p > 1 and diverges

when p ≤ 1
∴ ∑u n is convergent if p > 1 and divergent if p ≤ 1 .

EXAMPLE 32
∞
(n + 1)x n ; x > 0
Test the convergence of the series ∑n =1 n3
SOLUTION

un =
( n + 1) x n ; u ( n + 2 ) x n+1
( n + 1)
n +1
n3 3

n + 2 n +1 ⎛ n + 2 ⎞⎛ n ⎞
3
un +1 n3
= x =⎜ .x
( n + 1) x ⎝ n + 1 ⎟⎠ ⎜⎝ n + 1 ⎟⎠
. .
un ( n + 1) n
3

⎛ 2⎞
un +1 ⎜ 1+ n ⎟ 1
Lt = Lt ⎜ ⎟ .x = x
n →∞ u 1 3
n
n →∞
⎜ 1+ ⎟ 1+⎛ 1 ⎞
⎝ n ⎠ ⎜⎝ n ⎟⎠
∴ By ratio test, ∑u n converges when x < 1 and diverges when x > 1 .
n +1
When x = 1, un =
n3
1
Take vn = ; By comparison test ∑ un is convergent ( give proof )
n2
∴ ∑ un is convergent if x ≤ 1 and divergent if x > 1.
Sequences and Series 31

EXAMPLE 33
Test the convergence of the series (JNTU 2002)
∞
⎛n 2
1 ⎞ 2.5.8 2.5.8.11 1 1.2 1.2.3
(i) ∑⎜ 2 n
+ 2 ⎟
n ⎠
(ii) 1 + + + ... (iii) + + +
n −1 ⎝ 1.5.9 1.5.9.13 3 3.5 3.5.7
SOLUTION
∞
⎛ n2 1 ⎞ ∞
n2 ∞ 1 n2 1
(i) ∑ ⎜ n+ 2⎟ =
n −1 ⎝ 2 n ⎠
∑
n =1 2
n
+∑ 2
n =1 n
Let un =
2 n
; vn = 2
n
( n + 1) ( n + 1) . 2n
2 2 2
u u 1 ⎛ 1⎞ 1
un +1 = ; n +1 = Lt n +1 = Lt . ⎜1 + ⎟ = < 1
2n +1 un 2n +1 n 2 n →∞ u
n
n →∞ 2
⎝ n⎠ 2

∴ By ratio test∑ u is convergent. By p –series test, ∑ v

n n is convergent.
∴ The given series ( ∑ u + ∑ v ) is convergent.n n

2.5.8 2.5.8.11
(ii) Neglecting the first term, the series can be taken as, + +
1.5.9 1.5.9.13
Here, 1st term has 3 fractions ,2nd term has 4 fractions and so on .
∴ nth term contains ( n + 2 ) fractions
2. 5. 8.......are in A. P.
∴ ( n + 2)
th
term = 2 + ( n + 1 ) 3 = 3n + 5 ;
∴ 1. 5. 9,.......are in A. P.
∴ ( n + 2 ) term = 1 + ( n + 1 ) 4 = 4n + 5
th

2.5.8..... ( 3n + 5 )
∴ un =
1.5.9..... ( 4n + 5 )
2.5.8..... ( 3n + 5 )( 3n + 8 )
un +1 =
1.5.9..... ( 4n + 5 )( 4n + 9 )
⎛ 8⎞
n⎜3+ ⎟
un +1
=
( 3n + 8) ; u
Lt n +1 = Lt ⎝
n⎠ 3
= <1
un ( 4n + 9 ) n →∞ u
n
n →∞ ⎛ 9⎞ 4
n⎜4 + ⎟
⎝ n⎠
∴ By ratio test, ∑u n is convergent.
32 Engineering Mathematics - I

(iii) 1, 2, 3, ........ are in A. P nth term = n ; 3. 5. 7..........are in A.P. nth term = 2n + 1

⎡ 1.2.3.....n ⎤
∴ un = ⎢ ⎥
⎣ 3.5.7..... ( 2n + 1) ⎦
⎡ 1.2.3.....n ( n + 1) ⎤
un +1 = ⎢ ⎥
⎣ 3.5.7..... ( 2n + 1)( 2n + 3) ⎦
un +1 ⎛ n + 1 ⎞
=⎜ ⎟
un ⎝ 2n + 3 ⎠
⎛ 1⎞
n. ⎜1 + ⎟
u n⎠ 1
Lt n +1 = Lt ⎝ = <1
n →∞ u n →∞ ⎛ 3⎞ 2
n
n⎜2 + ⎟
⎝ n⎠
∴ By ratio test, ∑ un is convergent.

EXAMPLE 34
∞
1.3.5.... ( 2n − 1) n −1
Test for convergence ∑n =1 2.4.6....2n
.x ( x > 0 ) (JNTU 2001)

SOLUTION
1.3.5.... ( 2n − 1) n −1
The given series of +ve terms has un = .x
2.4.6....2n
1.3.5.... ( 2n + 1) n
and un +1 = x
2.4.6.... ( 2n + 2 )

un +1
Lt = Lt ⎜
⎛ 2n + 1 ⎞
x Lt
2n 1 + 1 (
2n .x = x )
⎟
( )
=
n →∞ u n →∞ 2n + 2
n ⎝ ⎠ n →∞
2n 1 + 2
2n
∴ By ratio test, ∑ un is converges when x < 1 and diverges when x > 1 when x = 1, the

test fails.
1.3.5.... ( 2n − 1)
Then un = < 1 and Lt un ≠ 0
2.4.6.....2n n →∞

∴ ∑ un is divergent. Hence ∑ un is convergent when x < 1, and divergent when x ≥ 1

Sequences and Series 33

EXAMPLE 35
2 6 2 ⎛ 2n − 2 ⎞ n −1
Test for the convergence of 1 + x + x + ........ + ⎜ n ⎟ x + ..... ( x > 0 )
5 9 ⎝ 2 +1 ⎠
(JNTU 2003)
SOLUTION
⎛ 2n − 2 ⎞ n −1
Omitting 1st term, un = ⎜ ⎟ x , ( n ≥ 2 ) and ' un ' are all +ve.
⎝ 2 +1 ⎠
n

=
(2 n +1
− 2) ⎛ un +1 ⎞
=
⎛ 2n +1 − 2 ⎞ ⎛ 2n + 1 ⎞
⎟×⎜ n
n
un +1 x ; Lt ⎜ ⎟ Lt . ⎜ n +1 ⎟ .x
(2 n +1
+ 1) n →∞
⎝ un ⎠ n→∞ ⎝ 2 + 1 ⎠ ⎝ 2 − 2 ⎠

= Lt
⎡ 2n +1 1 − 1
⎢ 2 n (
2n 1 + 1 n ⎤
2 .x ⎥ = x ; ) ( )
( ) ( )
.
n →∞ ⎢ n +1 ⎥
⎢⎣ 2 1 + 2n +1 2 1 − 2n ⎥⎦
1 n 2

Hence, by ratio test, ∑ un converges if x < 1 and diverges if x > 1.

2n − 2
When x = 1, the test fails. Then un = ; Lt un = 1 ≠ 0 ; ∴
2n + 1 n→∞
∑u n diverges

Hence ∑u n is convergent when x < 1 and divergent x > 1

EXAMPLE 36
( 3 − 4i )
n
∞
Using ratio test show that the series ∑
n=0 n!
converges (JNTU 2000)

SOLUTION
( 3 − 4i ) ( 3 − 4i )
n n +1
⎛ un +1 ⎞ ⎛ 3 − 4i ⎞
un = ; un +1 = ; Lt ⎜ ⎟ = nLt ⎜ ⎟ = 0 <1
n! (n + 1)!
⎝ un ⎠ →∞ ⎝ n + 1 ⎠
n →∞

Hence, by ratio test, ∑u n converges.

EXAMPLE 37
2 3 2 4 3
Discuss the nature of the series, x+ x + x + ........∞ ( x > 0 ) (JNTU 2003)
3.4 4.5 5.6
SOLUTION
Since x > 0 , the series is of +ve terms ;
34 Engineering Mathematics - I

un =
( n + 1) x n > u = ( n + 2 ) x n+1
( n + 2 )( n + 3) n +1
( n + 3)( n + 4 )
⎡ ⎤
un +1 ⎡ ( n + 2 ) .x ⎤ ⎢ n (1 + 2 n ) .x
2 2 2
⎥
Lt =⎢ ⎥ = Lt ⎢ ⎥ = x;
n →∞ u
n ⎢
⎣ ( n + 1)( n + 4 ) (
⎥⎦ n→∞ ⎢ n 2 1 + 5 + 4 2
⎣ n n ) ⎥⎦
Therefore by ratio test, ∑ un converges if x < 1 and diverges if x >1

When x = 1, the test fails; Then un =

( n + 1) ;
( n + 2 )( n + 3)
1 u
Taking vn = ; Lt = n = 1 ≠ 0
n n →∞ vn
∴ By comparison test ∑u n and ∑v n behave same way. But ∑v n is divergent by p-

series test. (p = 1);

∴∑ un is diverges when x = 1
∴ ∑u n is convergent when x < 1 and divergent when x ≥ 1

EXAMPLE 38
3.6.9.....3n.5n
Discuss the nature of the series ∑ 4.7.10.....( 3n + 1)( 3n + 2 ) (JNTU 2003)

SOLUTION
3.6.9.....3n 5n
Here, un = ;
4.7.10..... ( 3n + 1) ( 3n + 2 )
3.6.9.....3n ( 3n + 3) 5n +1
un +1 = ;
4.7.10..... ( 3n + 1)( 3n + 4 )( 3n + 5 )

Lt
un +1
= Lt
( 3n + 2 )( 3n + 3) .5
n →∞ u n →∞ ( 3n + 4 )( 3n + 5 )
n

= Lt
⎡ 5.9n 2 1 + 2
⎢ 3n (1+ 3
3n )( ) ⎤⎥ = 5 > 1
n →∞ ⎢
⎣
9n 1 + 4
2
(
3n
1+ 5
3n )( ) ⎥⎦
∴ By ratio test, ∑ un is divergent.
Sequences and Series 35

EXAMPLE 39
∞
Test for convergence the series ∑n
n =1
1− n

SOLUTION
un = n1− n ; un +1 = ( n + 1) ;
−n

u n +1 (n + 1)– n
n
nn 1⎛ n ⎞
= = = ⎜ ⎟
un n 1– n
n(n + 1) n
n ⎝ n +1⎠
n
u 1 ⎛ 1 ⎞⎟ 1
Lt n +1 = Lt . ⎜ = 0. = 0 < 1
n →∞ n ⎜
n →∞ u
1+ 1 ⎟ e
n
⎝ n⎠
∴ By ratio test ∑ un , is convergent

EXAMPLE 40
∞
2n3
Test the series ∑ , for convergence.
n =1 n

SOLUTION
2 ( n + 1)
3
2n3
un = ; un +1 =
n n +1

( )
2

un +1 2 ( n + 1)
3
n ( n + 1)
2
1+ 1
= × 3= = n ;
un n +1 2n n 3
n
u
Lt n +1 = 0 < 1 ;
n →∞ u
n

∴ By ratio test, ∑u n is convergent.

EXAMPLE 41
2n n !
Test convergence of the series ∑ nn
SOLUTION
2n n ! 2n +1 ( n + 1) !
un = ; u = ;
( n + 1)
n +1 n +1
nn
36 Engineering Mathematics - I

un +1 2 ( n + 1) ! n n
n +1 n
⎛ n ⎞
= . n = 2⎜ ⎟
( n + 1)
n +1
un 2 n! ⎝ n +1 ⎠
un +1 1 2
Lt = 2 Lt = < 1 (since 2 < e < 3)
( )
n →∞ u n
n
n →∞
1+ 1 e
n
∴ By ratio test, ∑u n is convergent.

EXAMPLE 42
Test the convergence of the series ∑u n where un is

n2 + 1 x n −1
2
⎛ 1.2.3....n ⎞
(a) (b) , ( a > 0) (c) ⎜ ⎟
3n + 1 ( 2n + 1) ⎝ 4.7.10.....3n + 3 ⎠
a

1 + 2n ⎛ 3n3 + 7 n 2 ⎞ n
(d) (e) ⎜ ⎟x
1 + 3n ⎝ 5 n 9
+ 11 ⎠
SOLUTION
⎛ un +1 ⎞ ⎡ ( n + 1)2 + 1 3n + 1 ⎤
(a) Lt ⎜ ⎟ = Lt ⎢ n +1 × 2 ⎥
n →∞
⎝ un ⎠ n→∞ ⎢⎣ 3 + 1 n + 1 ⎥⎦
⎡ n 2 ⎛1 + 2 + 2 ⎞ 3 n ⎛1 + 1 ⎞⎟ ⎤
Lt ⎢ ⎜⎝ n ⎟
n2 ⎠
⎜
⎝ 3n ⎠ ⎥
= ⎢ . ⎥
n → ∞ ⎢ n 2 ⎛1 + 1 ⎞ n +1 ⎛ ⎞
3 ⎜1 + n +1 ⎟ ⎥
⎜ ⎟ 1
⎣ ⎝ n2 ⎠ ⎝ 3 ⎠⎦
1
= <1
3
∴ By ratio test, ∑u n is convergent.
⎡ xn ( 2n + 1) ⎤
a
⎛ un +1 ⎞
(b) Lt ⎜ ⎟ = Lt ⎢ × ⎥
⎝ un ⎠ n →∞ ⎣⎢ ( 2n + 3) x n −1 ⎥⎦
n →∞ a

= Lt
⎡ 2a n a 1 + 1 a ⎤
⎢ (
2n . x ⎥ = x )
⎢ ⎥
( )
n →∞ a
⎢⎣ 2 n 1 + 2n
a a 3
⎥⎦

By ratio test, ∑u n convergence if x < 1 and diverges if x > 1.

Sequences and Series 37

1 1
When x = 1, the test fails; Then, un = ; Taking vn = we have,
( 2n + 1)
a
na
a
⎛u ⎞ ⎛ n ⎞ 1 1
Lt ⎜ n ⎟ = nLt ⎜ ⎟ = nLt = ≠ 0 and finite ( since a > 0 ).
→∞ 2n + 1
( )
a
n →∞ v
⎝ n ⎠ ⎝ ⎠ →∞
2+ 1 2a
n
∴ By comparison test, ∑u n and ∑v n have same property

But p –series test, we have

(i) ∑v n convergent when a > 1

and (ii) divergent when a ≤ 1

∴ To sum up, (i) x < 1, ∑u n is convergent ∀a .

(ii) x > 1, ∑u n is divergent ∀a .

(iii) x = 1, a > 1, ∑u n is convergent, and

(iv) x = 1, a ≤ 1 , ∑ u n is divergent.

1.2.3....n ( n + 1) 4.7.10.... ( 3n + 3) ⎤
2
u ⎡
(c) Lt n +1 = Lt ⎢ × ⎥
n →∞ u n →∞ 4.7.10.... ( 3n + 3 )( 3n + 6 ) 1.2.3....n
n ⎣ ⎦
⎡ ( n + 1) ⎤
2
1
= Lt ⎢ ⎥ = <1 ;
n →∞ 3 ( n + 2 )
⎣⎢ ⎦⎥ 9
∴ By ratio test, ∑u n is convergent

⎡ (1 + 2n +1 ) (1 + 3n ) ⎤
1
2
un +1
(d) Lt = Lt ⎢ × ⎥
⎢⎣ (1 + 3 ) (1 + 2 ) ⎥⎦
n →∞ u n →∞ n +1 n
n

1
⎡ n +1 ⎛ 1 ⎞ n⎛ 1 ⎞⎤ 2

⎢ 2 ⎜1 + 2n +1 ⎟ 3 ⎜1 + 3n ⎟ ⎥ ⎛2⎞
1

= Lt ⎢ ⎝ ⎠× ⎝ ⎠⎥ =⎜ ⎟
2
<1
⎛
⎢ 3n +1 1 + ⎞ ⎛ ⎞ ⎝3⎠
2n ⎜1 + n ⎟ ⎥
n →∞ 1 1
⎢⎣ ⎜ ⎟
⎝ 2 ⎠ ⎥⎦
n +
⎝ 3 ⎠
1

∴ By ratio test, ∑u n is convergent.

38 Engineering Mathematics - I

un +1 ⎡ 3 ( n + 1)3 + 7 ( n + 1)2 5n9 + 11 ⎤

(e) Lt = Lt ⎢ × 3 × x⎥
n →∞ u
( + ) + n +
9
n
n →∞
⎢
⎣ 5 n 1 11 3 7 ⎥⎦

= Lt ⎢
⎡ 3
(
3

⎢ 3n 1 + n + 7 n 1 + n
1 2 1 ) ( )
2

×
( 5 n )
5n9 1 + 11 ⎤
⎥
× x⎥
9

3n )
( ) 3n (1 + 7
n →∞ 9
5n 9 1 + 1 + 11
3
⎢ n ⎥
⎣ ⎦
3

⎡ 3⎧
(
1 + 7 1 + 1 ⎫⎬ 5n9 1 + 11) ⎤
( ) ( )
3 2

⎢ 3n ⎨ 1 + ⎥
⎩ n 3n n ⎭× 5n × x ⎥ = x
9
= Lt ⎢
n →∞
⎢
⎢⎣
⎧
5n 9 ⎨ 1 + 1
⎩ n
9 11 ⎫
+ 9⎬
5 n ⎭
( 3)
3n3 1 + 7 3
n
⎥
⎥⎦
( )
∴ By ratio test, ∑ un converges when x < 1 and diverges when x > 1.

When x = 1, the test fails,

un =
3n3 1 + 7( = 6
3 1 + 3n
7
3n ) ( )
( ) ( )
Then
5n9 1 + 11 9 5n 1 + 11
5n 5n9
1 u 3
Taking vn = 6 , we observe that Lt n = ≠ 0
n n →∞ vn 5
∴ By comparison test and p series test, we conclude that ∑u n is convergent.
∴ ∑u n is convergent when x ≤ 1 and divergent when x > 1.

Exercise – 1.2
1. Test the convergency or divergency of the series whose general term is :
xn
(a) ............................... [Ans : x < 1cgt , x ≥ 1dgt ]
n
(b) nx n −1 ........................... [Ans : x < 1cgt , x ≥ 1dgt ]
⎛ 2n − 2 ⎞ n −1
(c) ⎜ n ⎟ x .............. [Ans : x < 1cgt , x ≥ 1dgt ]
⎝ 2 +1 ⎠
⎛ n2 + 1 ⎞ n
(d) ⎜ 2 ⎟ x ................... [Ans : x < 1cgt , x ≥ 1dgt ]
⎝ n −1 ⎠
n
(e) ........................ [Ans: cgt.]
nn
Sequences and Series 39

4n. n
(f) .......................... [Ans: dgt.]
nn
( n + 1)
3 n

(g) ....................... [Ans: cgt.]

( 3 + 1)
n

2. Determine whether the following series are convergent or divergent :

x x2 x3
(a) + + + .............. [Ans : x ≤ 1cgt , x > 1dgt ]
1.2 3.4 5.6
x x 2 x3
(b) 1 + 2 + 2 + 2 + .............. [Ans : x ≤ 1cgt , x > 1dgt ]
2 3 4
1 x x2
(c) + + + ...... [Ans : x ≤ 1cgt , x > 1dgt ]
1.2.3 4.5.6 7.8.9
x x 2 x3 xn
(d) 1 + + + + ..... 2 + .... [Ans : x ≤ 1cgt , x > 1dgt ]
2 5 10 n +1
1.2 2.3 3.4
(e) + 2 + 3 + ............... [Ans : x > 1cgt , x ≤ 1dgt ]
x x x

1.3.4 Raabe’s Test

⎧⎪ ⎛ un ⎞ ⎫⎪
Let ∑u n be series of +ve terms and let Lt ⎨n ⎜ − 1⎟ ⎬ = k
n →∞
⎪⎩ ⎝ un +1 ⎠ ⎪⎭
Then
(i) If k > 1, ∑u n is convergent. (ii) If k < 1, ∑u n is divergent. (The test fails if k = 1)
1
Proof : Consider the series ∑v = ∑ n
n p

⎡ vn ⎤ ⎡⎛ n + 1 ⎞ p ⎤ ⎡⎛ 1 ⎞ p ⎤
n⎢ − 1⎥ = n ⎢⎜ ⎟ − 1⎥ = n ⎢⎜1 + ⎟ − 1⎥
⎣ vn +1 ⎦ ⎢⎣⎝ n ⎠ ⎥⎦ ⎢⎣⎝ n ⎠ ⎥⎦
⎡⎛ p p ( p − 1) 1 ⎞ ⎤
= n ⎢⎜ 1 + + . 2 + .... ⎟ − 1⎥
⎢⎣⎝ n 2 n ⎠ ⎥⎦
p ( p − 1) 1 ⎧v ⎫
= p+ . + .......... Lt n ⎨ n − 1⎬ = p
2 n n →∞
⎩ vn +1 ⎭
40 Engineering Mathematics - I

⎧ un ⎫
Case (i) In this case, Lt n ⎨ − 1⎬ = k > 1
n →∞
⎩ un +1 ⎭
We choose a number ‘p’ ∋ k > p > 1 ; Comparing the series ∑u n with

∑v n which is convergent , we get that ∑u n will converge if after some

fixed number of terms
p
un v ⎛ n +1⎞
> n =⎜ ⎟
un +1 vn +1 ⎝ n ⎠
⎛ u ⎞ p ( p − 1) 1
i.e. If, n ⎜ n − 1⎟ > p + . + .........from (1)
u
⎝ n +1 ⎠ ∠ 2 n
⎛ u ⎞
i.e., If Lt n ⎜ n − 1⎟ > p
n →∞
⎝ un +1 ⎠
i.e., If k > p, which is true . Hence ∑u n is convergent .The second case also
can be proved similarly.

Solved Examples
EXAMPLE 43
Test for convergence the series
1 x3 1.3 x5 1.3.5 x 7
x+ . + . + . + ..... (JNTU 2006, 2008)
2 3 2.4 5 2.4.6 7
SOLUTION
Neglecting the first tem ,the series can be taken as ,
1 x3 1.3 x5 1.3.5 x 7
. + . + . + .....
2 3 2.4 5 2.4.6 7
1.3.5....are in A.P. nth term = 1 + ( n − 1) 2 = 2n − 1

2.4.6...are in A.p. nth term = 2 + ( n − 1) 2 = 2n

3.5.7.....are in A.P nth term = 3 + ( n − 1) 2 = 2n + 1

1.3.5.... ( 2n − 1) x 2 n +1
∴ un ( nth term of the series) = .
2.4.6.... ( 2n ) 2n + 1
Sequences and Series 41

1.3.5.... ( 2n − 1)( 2n + 1) x 2 n +3
un +1 = .
2.4.6.... ( 2n )( 2n + 2 ) 2n + 3

un +1 1.3.5.... ( 2n + 1) x 2 n +3 2.4.6....2n ( 2n + 1)
= . . .
un 2.4.6.... ( 2n + 2 ) ( 2n + 3) 1.3.5.... ( 2n − 1) x 2 n +1

( 2n + 1) x 2
2

=
( 2n + 2 )( 2n + 3)
2
⎛ 1 ⎞
4n ⎜ 1 + ⎟
2
u ⎝ 2n ⎠
∴ Lt n +1 = Lt x2 = x2
n →∞ u n →∞ ⎛ 2 ⎞ ⎛ 3 ⎞
n
4n 2 ⎜ 1 + ⎟ ⎜ 1 + ⎟
⎝ 2n ⎠ ⎝ 2n ⎠
∴ By ratio test, ∑ un converges if x < 1 and diverges if x > 1
If x = 1 the test fails.

Then x2 = 1 and
un
=
( 2n + 2 )( 2n + 3)
un +1 ( 2n + 1)
2

un
−1 =
( 2n + 2 )( 2n + 3) − 1 = 6n + 5
un +1 ( 2n + 1) ( 2n + 1)
2 2

⎧⎪ ⎛ un ⎞ ⎫⎪ ⎛ 6n 2 + 5n ⎞
Lt ⎨n ⎜ − 1⎟ ⎬ = Lt ⎜ 2 ⎟
n →∞ 4n + 4n + 1
⎩⎪ ⎝ un +1 ⎠ ⎭⎪ ⎝ ⎠
n →∞

⎛ 5⎞
n2 ⎜ 6 + ⎟
= Lt ⎝ n⎠ 3
= >1
n →∞ ⎛ 4 1 ⎞ 2
n2 ⎜ 4 + + 2 ⎟
⎝ n n ⎠
By Raabe’s test, ∑ un converges. Hence the given series is convergent when x ≤ 1 an
divergent when x > 1 .

EXAMPLE 44
Test for the convergence of the series (JNTU 2007)
3 3.6 2 3.6.9 3
1+ x+ x + x + .....; x > 0
7 7.10 7.10.13
42 Engineering Mathematics - I

SOLUTION
Neglecting the first term,
3.6.9....3n
un = .x n
7.10.13....3n + 4
3.6.9....3n ( 3n + 3)
un +1 = .x n +1
7.10.13.... ( 3n + 4 )( 3n + 7 )
un +1 3n + 3 u
= .x ; Lt n +1 = x
un 3n + 7 n →∞ u
n

∴ By ratio test, ∑u n is convergent when x < 1 and divergent when x > 1.

When x = 1 The ratio test fails. Then

un 3n + 7 un 4
= ; −1 =
un +1 3n + 3 un +1 3n + 3

⎧⎪ ⎛ u ⎞ ⎫⎪ ⎛ 4n ⎞ 4
Lt ⎨n ⎜ n − 1⎟ ⎬ = Lt ⎜ ⎟ = >1
n →∞ 3n + 3
⎩⎪ ⎝ un +1 ⎠ ⎭⎪ ⎝ ⎠ 3
n →∞

∴ By Raabe’s test, ∑u n is convergent .Hence the given series converges if x ≤ 1 and

diverges if x > 1.

EXAMPLE 45

12.52.92.... ( 4n − 3)
2
∞
Examine the convergence of the series ∑
n =1 42.82.122.... ( 4n )
2

SOLUTION
12.52.92.... ( 4n − 3) 12.52.92.... ( 4n − 3) ( 4n + 1)
2 2 2

un = ; un +1 =
42.82.122.... ( 4n ) 42.82.122.... ( 4n ) ( 4n + 4 )
2 2 2

( 4n + 1) = 1 (verify)
2
u
Lt n +1 = Lt
n →∞ u
( 4n + 4 )
n →∞ 2
n

∴ The ratio test fails. Hence by Raabe’s test, ∑u n is convergent. (give proof)
Sequences and Series 43

EXAMPLE 46
( n)
2

Find the nature of the series ∑ 2n

xn , ( x > 0) (JNTU 2003)

SOLUTION
( n) ( n + 1)
2 2

un = n
.x ; un +1 = .x n +1
2n 2n + 2
( n + 1)
2
un +1
= x;
un ( 2n + 1)( 2n + 2 )
( )
2

un +1 n2 1 + 1 x
Lt = Lt n .x =
n →∞ u
n
n →∞
4n 2 1 + 1
2n (
1+ 2
2n
4 )( )
x
∴ By ratio test, ∑u n converges when
4
< 1 , i. e ; x < 4; and diverges when x >4;

When x = 4, the test fails.

un
=
( 2n + 1)( 2n + 2 )
un +1 4 ( n + 1)
2

un −2n − 2 −1 ⎡ ⎛ u ⎞ ⎤ −1
−1 = = ; Lt ⎢ n ⎜ n − 1⎟ ⎥ = <1
un +1 4 ( n + 1)
2
2 ( n + 1) n →∞
⎣ ⎝ u n +1 ⎠ ⎦ 2
∴ By ratio test, ∑u n is divergent
Hence ∑u n is convergent when x < 4 and divergent when x > 4

EXAMPLE 47
4.7.... ( 3n + 1) n
Test for convergence of the series ∑ 1.2.3....n
x (JNTU 1996)

SOLUTION
4.7.... ( 3n + 1) n 4.7.... ( 3n + 1)( 3n + 4 ) n +1
un = x ; un +1 = x
1.2.3....n 1.2.3....n ( n + 1) .
un +1 ⎡ ( 3n + 4 ) ⎤
Lt = Lt ⎢ .x ⎥ = 3 x
n →∞ u
n
n →∞
⎣ ( n + 1) ⎦
44 Engineering Mathematics - I

1 1
∴ By ratio test ∑u n converges if 3 x < 1 i.e., x <
3
and diverges if x > ;
3
1
If x = , the test fails
3
1 ⎡u ⎤ ⎡ (n + 1)3 ⎤ ⎡ −1 ⎤ 1
When x = , n ⎢ n – 1⎥ = n ⎢ – 1⎥ = n ⎢ ⎥ =−
3 ⎣ u n +1 ⎦ ⎣ 3n + 4 ⎦ ⎣ 3n + 4 ⎦ ⎛ 4⎞
⎜3+ ⎟
⎝ n⎠
⎡u ⎤ 1
Lt n ⎢ n − 1⎥ = − < 1
n →∞
⎣ un +1 ⎦ 3
∴ By Raabe’s test, ∑u n is divergent.
1 1
∴ ∑u n is convergent when x <
3
and divergent when x ≥
3
EXAMPLE 48
3x 4 x 2 5 x3
Test for convergence 2 + + + + ........... ( x > 0 ) (JNTU 2003)
2 3 4
SOLUTION

The nth term un =

( n + 1) x n−1 ; un +1 =
( n + 2) xn ; un +1 n ( n + 2 )
= .x
n ( n + 1) un ( n + 1)
2

Lt
un +1
= Lt
n2 1 + 2 (
n .x = x )
n →∞ u
( )
n →∞ 2 2
n n 1+ 1
n
∴ By ratio test, ∑ un is convergent if x < 1 and divergent if x > 1
If x = 1, the test fails.
⎡ un ⎤ ⎡ ( n + 1)2 ⎤ ⎡ 1 ⎤
Then Lt n ⎢ − 1⎥ = Lt n ⎢ − 1⎥ = Lt n ⎢ ⎥ = 0 <1
⎣ un +1 ⎦ n→∞ ⎢⎣ n ( n + 2 ) ⎥⎦ ⎣ n ( n + 2) ⎦
n →∞ n →∞

∴ By Raabe’s test ∑u n is divergent

∴ ∑u n is convergent when x < 1 and divergent when x ≥ 1

EXAMPLE 49
3 3.6 3.6.9
Find the nature of the series + + + ......∞ (JNTU 2003)
4 4.7 4.7.10
Sequences and Series 45

SOLUTION
3.6.9.....3n 3.6.9.....3n ( 3n + 3)
un = ; un +1 =
4.7.10..... ( 3n + 1) 4.7.10..... ( 3n + 1)( 3n + 4 )

un +1 3n + 3
= ; Lt
un +1
= Lt
3n 1 + 3 (
3n = 1 )
un 3n + 4 n →∞ un n →∞
3n 1 + 4 (
3n )
Ratio test fails.
⎡ ⎧u ⎫⎤ ⎡ ⎛ 3n + 4 ⎞ ⎤
∴ Lt ⎢ n ⎨ n − 1⎬⎥ = Lt ⎢ n ⎜ − 1⎟ ⎥
n →∞
⎣ ⎩ un +1 ⎭⎦
n →∞
⎣ ⎝ 3n + 3 ⎠ ⎦
n n 1
= Lt = Lt = <1
n →∞ 3 ( n + 1) n →∞
3n 1 + 1(n
3 )
∴ By Raabe’s test ∑ un is divergent.

EXAMPLE 50
If p, q > 0 and the series
1 p 1.3. p ( p + 1) 1.3.5 p ( p + 1)( p + 2 )
1+ + + + ....
2 q 2.4.q ( q + 1) 2.4.6 q ( q + 1)( q + 2 )
is convergent , find the relation to be satisfied by p and q.
SOLUTION
1.3.5..... ( 2n − 1) p ( p + 1) ..... ( p + n − 1)
un = [neglecting 1st term]
2.4.6.....2n q ( q + 1) ..... ( q + n − 1)
1.3.5..... ( 2n − 1)( 2n + 1) p ( p + 1) ..... ( p + n − 1)( p + n )
un +1 =
2.4.6.....2n ( 2n + 2 ) q ( q + 1) ..... ( q + n − 1)( q + n )
un +1 ( 2n + 1) ( p + n )
= ;
un ( 2n + 2 ) ( q + n )

Lt
un +1
= Lt ⎢
⎡
(
⎢ 2 n 1 + 2n
1 )
n 1+ p (
n ) ⎤⎥⎥ = 1
( ) ( ) ⎥⎦
.
n →∞ u q
⎢ 2 n 1 + 2n n 1 + n
n →∞ 1
n
⎣
∴ ratio test fails.
Let us apply Raabe’s test
46 Engineering Mathematics - I

⎡ ⎛ u ⎞⎤ ⎡ ⎧⎪ ( q + n )( 2n + 2 ) ⎫⎪⎤
Lt ⎢ n ⎜ n − 1⎟ ⎥ = Lt ⎢ n ⎨ − 1⎬⎥
n →∞ u
⎣ ⎝ n +1 ⎠ ⎦
n →∞
⎪
⎣⎢ ⎩ ( p + n )( 2 n + 1) ⎭⎪⎦⎥
⎡ ⎧ ⎫⎤
⎢ ⎪ 2q ( n + 1) − p ( 2n + 1) + n ⎪⎥
Lt ⎢ n ⎨
( )(
⎬⎥
n →∞
⎢ ⎪ n2 1 + p
⎣ ⎩ n
2+ 1
n ⎭⎦ )
⎪⎥

Lt ⎢ n ( ) (
⎡ 2q 1 + 1 − p 2 + 1 + 1 ⎤
n )
⎥ = 2q − 2 p + 1
n →∞ ⎢ 2 ⎥ 2
⎣ ⎦
2q − 2 p + 1
Since ∑ un is convergent, by Raabe’s test, >1
2
⇒ q − p > 1 , is the required relation.
2

Exercise 1.3
∞
1. Test whether the series ∑u
1
n is convergent or divergent where

22.42.62..... ( 2n − 2 )
2

un = .x 2 n [Ans : x ≤ 1cgt , x > 1dgt ]

3.4.5..... ( 2n − 1)( 2n )
2. Test for the convergence the series
∞
4.7.10..... ( 3n + 1) n 1 1
∑1 n
x [Ans : x <
3
cgt , x ≥ dgt ]
3
3. Test for the convergence the series :
22.42 22.42.52.7 2 22.42.52.7 2.82.102
(i) + + + ... [Ans : divergent]
32.32 32.32.62.62 32.32.62.62.92.92

3.4 4.5 2 5.6 3

(ii) x+ x + x + ..... ( x > 0 ) [Ans : cgt if x ≤ 1 dgt if , x > 1]
1.2 2.3 3.4

1.3.5..... ( 2n − 1) xn
(iii) ∑ 2.4.6.....2n ( 2n + 2 ) ( x > 0 ) [Ans : cgt if x ≤ 1 dgt if , x > 1]
.
Sequences and Series 47

( 1) ( 2) ( 3)
2 2 2
x2 x3
(iv) 1+ x+ + + ...... ( x > 0 )
2 4 6
[Ans : cgt if x < 4 and dgt if , x ≥ 4 ]

1.3.5 Cauchy’s Root Test

∑u ∑u
1
Let n be a series of +ve terms and let Lt un n
= l . Then n is convergent when
n →∞

l < 1 and divergent when l > 1

= l < 1 ⇒ ∃a +ve number ' λ ' ( l < λ < 1) ∋ un

1 1
Proof : (i) Lt un n n
< λ , ∀n > m
n →∞

(or) un < λ n , ∀n > m

Since λ < 1, ∑ λ n is a geometric series with common ratio < 1 and
therefore convergent.
Hence ∑
un is convergent.
1
(ii) Lt un n
= l >1
n →∞
1
∴ By the definition of a limit we can find a number r ∋ un n
> 1∀n > r
i.e., un > ∀n > r
i.e., after the 1st ‘r ‘ terms , each term is > 1.
Lt
n →∞
∑
un = ∞ ∴ un is divergent. ∑
Note : When Lt ( u
n →∞
n
1
n
) = 1, the root test can’t decide the nature of ∑ u n . The fact of

this statement can be observed by the following two examples.

1 3
⎛ 1⎞ n ⎛ 1 ⎞
∑
1
1. Consider the series 1 : − Ltun n
= Lt ⎜ 3 ⎟ = Lt ⎜ 1 ⎟ = 1
n3 n →∞ n →∞ ⎝ n ⎠ n →∞ ⎝ n n ⎠

1
∑ 1 n , in which Ltu
1
2. Consider the series n
n
= Lt 1 = 1
n →∞ n →∞ n n
1
In both the examples given above, Ltu
n →∞
n
n
= 1 . But series (1) is convergent

(p-series test)
And series (2) is divergent. Hence when the limit=1, the test fails.
48 Engineering Mathematics - I

Solved Examples
EXAMPLE 51
Test for convergence the infinite series whose nth terms are:
1 1 1
(i) (ii) (iii) (JNTU 1996, 1998, 2001)
n2n (log n) n ⎡ 1⎤
n2

⎢⎣1 + n ⎥⎦
SOLUTION

1 1 1
un = , un n = 2 ; Lt un n = Lt 2 = 0 < 1;
1 1
(i) 2n
n n n →∞ n →∞ n

By root test ∑ un is convergent.

1 1 1
un = ; un n = ; Lt un n = Lt = 0 < 1;
1 1
(ii) n
(log n) log n n →∞ n →∞ log n
∴ By root test , ∑ un is convergent.
1 1 1 1 1 1
(iii) un = ; un n
= n Ltu n
n
= Lt n
= < 1;
⎛ 1⎞
n2
⎛ 1⎞ n →∞ n →∞ ⎛ 1⎞ e
⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟
⎝ n⎠ ⎝ n⎠ ⎝ n⎠
∴ By root test ∑u n is convergent.

EXAMPLE 52
Find whether the following series are convergent or divergent.
⎡⎣( n + 1) x ⎤⎦
n
∞ ∞
1 1 1 1
(i) ∑ n (ii) 1 + 2 + 3 + 4 + ..... (iii) ∑
n =1 3 − 1 2 3 4 n =1 n n +1

SOLUTION
1
⎛ ⎞ n

1 ⎛ 1 ⎞
1
n ⎜ 1 ⎟
(i) un n
=⎜ n ⎟ =⎜ ⎟
⎝ 3 −1 ⎠ ⎜ 3n ⎛1 − 1 ⎞ ⎟
⎜ ⎜ 3n ⎟ ⎟
⎝ ⎝ ⎠⎠
Sequences and Series 49

1
⎛ ⎞ n

⎜ 1 ⎟ 1
∑u
1
Lt un n
= Lt ⎜ ⎟ = < 1 ; By root test, n is convergent.
n →∞ n →∞
⎜ n⎛ 1 ⎞⎟ 3
⎜ 3 ⎜1 − 3n ⎟⎟
⎝ ⎝ ⎠⎠
1
1 ⎛ 1 ⎞ n

∑u
1
(ii) un = n ; Lt un n = Lt ⎜ n ⎟ = 0 < 1 ; By root test, n is convergent.
n n →∞ n →∞
⎝n ⎠

⎡( n + 1) x ⎤⎦
n

(iii) un = ⎣
n n +1
1
1
⎡ {( n + 1) x}n ⎤ n

Lt un n
= Lt ⎢ ⎥
n →∞ n →∞ ⎢ n n +1 ⎥
⎣ ⎦
1
⎡ ⎧ ( n + 1) x ⎫n 1 ⎤ n ⎛ n +1⎞ 1
Lt ⎢ ⎨ ⎬ . ⎥ = nLt ⎜ ⎟ x. 1
n →∞
⎢⎣ ⎩ n ⎭ n ⎥⎦ →∞
⎝ n ⎠ n n
⎛ 1⎞ 1 1 ⎛ 1 ⎞
Lt ⎜ 1 + ⎟ x. 1 = Lt x. 1 = x ⎜ since Lt x. 1 = 1⎟
n →∞
⎝ n⎠ n n n →∞
n n ⎝ n →∞
n n ⎠
∴ ∑ un is convergent if x < 1 and divergent if x > 1 and when x = 1 the
test fails.
( n + 1)
n
1
Then un = n +1
; Take vn =
n n
un ( n + 1) ( n + 1) = ⎛1 + 1 ⎞ ;
n n n
un
= n +1
.n = ⎜ ⎟ Lt = e >1
vn n nn ⎝ n⎠ n →∞ v
n

∴ By comparison test, ∑ un is divergent.

∑v ( n diverges by p –series test )
Hence ∑ u n is convergent if x < 1 and divergent x ≥ 1

EXAMPLE 53
2
nn
If un = n2
, show that ∑u n is convergent.
( n + 1)
50 Engineering Mathematics - I

1
⎡ n2 ⎤ n
n
n nn ⎛ n ⎞
= Lt ⎢ ⎥
1
Lt un n
; = Lt = = Lt ⎜ ⎟
n →∞ ⎢ ⎥ n →∞ n + 1
( n + 1)
n
⎝ ⎠
n2

( )
n →∞ n →∞
⎣ n + 1 ⎦
n
⎛ ⎞
⎜ 1 ⎟ 1
= Lt ⎜ ⎟ = <1; ∴
e
∑u n converges by root test .
⎜ 1+ 1 ⎟
n →∞

⎝ n⎠
EXAMPLE 54
2 3
1 ⎛2⎞ ⎛3⎞
Establish the convergence of the series + ⎜ ⎟ + ⎜ ⎟ + ........
3 ⎝5⎠ ⎝7⎠
SOLUTION
n
⎛ n ⎞ 1 ⎛ n ⎞ 1
un = ⎜ ⎟ .........(verify); Lt un n
= Lt ⎜ ⎟ = <1
⎝ 2n + 1 ⎠ n →∞ n →∞ 2 n + 1
⎝ ⎠ 2
By root test, ∑ un is convergent.

EXAMPLE 55
∞
n n
Test for the convergence of ∑
n =1 n +1
.x

SOLUTION
1 1
⎛ ⎞2 ⎛ ⎞2
⎜ 1 ⎟ n 1 ⎜ 1 ⎟
un = ⎜ ⎟ .x ; Lt un n = Lt ⎜ ⎟ .x = x
1
⎜ 1+ ⎟
n →∞ n →∞ 1
⎜ 1+ ⎟
⎝ n⎠ ⎝ n⎠
∴ By root test, ∑ un is convergent if x < 1 and divergent if x > 1 .
n 1
When x = 1 : un = , taking vn = 0 and applying comparison test , it can be
n +1 n
seen that is divergent
∑ un is convergent if x < 1 and divergent if x ≥ 1 .

EXAMPLE 56

∑( )
∞ 1 n
Show that n n − 1 converges.
n =1
Sequences and Series 51

SOLUTION

( )
1 n
un = n n
−1

Lt un
n →∞
1
n
( 1
)
= Lt n n − 1 = 1 − 1 = 0 < 1 since Lt n
n →∞
( n →∞
1
n
=1 ;)
∴ ∑u n is convergent by root test.

EXAMPLE 57
n
th ⎛n+2⎞ n
Examine the convergence of the series whose n term is ⎜ ⎟ .x
⎝ n+3⎠
SOLUTION
n
⎛n+2⎞ n 1 ⎛n+2⎞
un = ⎜ ⎟ .x ; nLt un n = Lt ⎜ ⎟x = x
⎝ n+3⎠ →∞ n →∞
⎝ n+3⎠
∴ By root test, ∑ un converges when x < 1 and diverges when x > 1 .
n
⎛ 2⎞
n ⎜1 + ⎟
⎛n+2⎞ n⎠
When x = 1 : un = ⎜ ⎟ ; nLt un = Lt ⎝
⎝ n+3⎠
n
→∞ n →∞
⎛ 3⎞
⎜ 1 + ⎟
⎝ n⎠
e2 1
= = ≠ 0 and the terms are all +ve .
e3 e
∴ ∑ un is divergent . Hence ∑ un is convergent if x < 1 and divergent if x ≥ 1 .

EXAMPLE 58
Show that the series,
−1 −2 −3
⎡ 22 2 ⎤ ⎡ 33 3 ⎤ ⎡ 44 4 ⎤
−
⎢ 12 1 ⎥ + −
⎢ 23 2 ⎥ + ⎢ 34 − 3 ⎥ + ...... is convergent (JNTU 2002)
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−n −n
⎡ ( n + 1)n +1 n + 1 ⎤ ⎛ n +1⎞
−n
⎡⎛ n + 1 ⎞ n ⎤
un = ⎢ n +1
− ⎥ ;= ⎜ ⎟ ⎢⎜ ⎟ − 1⎥
⎢⎣ n n ⎥⎦ ⎝ n ⎠ ⎢⎣⎝ n ⎠ ⎥⎦
−n −1
⎛ 1⎞
−n
⎡⎛ 1 ⎞ n ⎤ 1 ⎛ 1 ⎞ ⎡⎛ 1 ⎞
−1 n
⎤
⎜1 + ⎟ ⎢⎜ + − ⎥ u = + ⎟ ⎢⎜ + ⎟ − 1⎥
n
1 ⎟ 1 ; n ⎜ 1 1
⎝ n⎠ ⎢⎣⎝ n ⎠ ⎥⎦ ⎝ n ⎠ ⎢⎣⎝ n ⎠ ⎥⎦
52 Engineering Mathematics - I

1 1
=
⎛ 1 ⎞ ⎪⎧⎛ 1 ⎞ n ⎪⎫
⎜1 + ⎟ 1 + − 1⎬
⎝ n ⎠ ⎨⎪⎜⎝ n ⎟⎠
⎩ ⎭⎪
1 1 1 1
∴ Lt un n = . = <1
n →∞ 1 e −1 e −1
∴ By root test, ∑ un is convergent.

EXAMPLE 59
∞
e− m
Test ∑u for convergence when um =
( )
m − m2
m =1
1+ 2
m
SOLUTION
1
⎡
( ) ⎤
m2 m

⎢ 1+ m
( )
2 m
1 ⎥ 1⎛ 2⎞ e2
Lt um m
= Lt ⎢ ⎥ ; Lt ⎜ 1 + ⎟ = = e >1
m →∞ m →∞ em m →∞ e
⎝ m⎠ e
⎢ ⎥
⎣ ⎦
Hence Cauchy’s root tells us that ∑u m is divergent.

EXAMPLE 60
n
Test the convergence of the series ∑e n2
.

SOLUTION
1
n n
∑u
1
Lt un n
= Lt n = 0 < 1 ∴ By root test, n is convergent.
n →∞ n →∞ e

EXAMPLE 61
( n + 1) .x + ......, x > 0
n n
2 32
Test the convergence of the series, 2 x + 3 x 2 + ...
1 2 n n +1
SOLUTION
1
1 ⎡ ( n + 1)n .x n ⎤ n
⎡⎛ n + 1 ⎞ 1 ⎤
Lt un n
= Lt ⎢ n +1 ⎥ = Lt ⎢⎜ ⎟ . 1 n .x ⎥
n →∞ n →∞
⎢⎣ n ⎥⎦ n →∞
⎣ ⎝ n ⎠ n ⎦
Sequences and Series 53

⎡⎛ 1 ⎞ 1 ⎤
= Lt ⎢⎜1 + ⎟ . 1 .x ⎥ = 1.1.x = x ⎡since Lt n n = 1⎤
1

⎢
⎣ ⎥⎦
n →∞
⎣⎝ n ⎠ n n ⎦ n →∞

∴ By root test, ∑ un converges if x < 1 and diverges when x > 1 .

When x = 1 , the test fails.
n
⎛ 1⎞ 1 1
Then un = ⎜ 1 + ⎟ . ; Take vn =
⎝ n⎠ n n
n
un ⎛ 1⎞
Lt = Lt ⎜1 + ⎟ = e ≠ 0
n →∞ v
n
n →∞
⎝ n⎠
∴ By comparison test and p-series test, ∑u n is divergent.
Hence ∑u n is convergent when x < 1 and divergent when x ≥ 1 .

Exercise 1.4
1. Test for convergence the infinite series whose nth terms are:
1
(a) ............................. [Ans : convergent]
2 −1n

1
(b) . ( n ≠ 1) ...................... [Ans : convergent]
( log )
2n

n
⎛ 3n + 1 ⎞ 4 4
(c) ⎜ .x ⎟ ................................. [Ans : x < cgt , x ≥ dgt ]
⎝ 4n + 3 ⎠ 3 3
xn
(d) ............................................ [Ans : cgt for all x ≥ 0 ]
n
n
(e) ............................................. [Ans : convergent]
nn
3n.∠n
(f) ........................................ [Ans : convergent]
n3
( 2n − 1)
2 n

(g) ................................... [Ans : convergent]

( 2n )
2n

( )
1 2n
(h) n n
−1 ................................... [Ans : convergent]
54 Engineering Mathematics - I

− n2
⎛ n −1 ⎞
(i) ⎜ ⎟ ................................... [Ans : divergent]
⎝ n ⎠
n
⎛ nx ⎞
(j) ⎜ ⎟ , ( x > 0 ) ......................... [Ans : x < 1 cgt , x ≥ 1 dgt]
⎝ n +1 ⎠
2. Examine the following series for convergence :
x x 2 x3
(a) 1 + + 2 + 3 + ....., x > 0 .................. [Ans : x ≤ 1cgt , x > 1dgt ]
2 3 4
2 3
1 ⎛2⎞ ⎛ 3 ⎞
(b) + ⎜ ⎟ + ⎜ ⎟ + ..... ....................... [Ans : convergent]
4 ⎝ 7 ⎠ ⎝ 10 ⎠

1.3.6 Integral Test

+ve term series,
φ (1) + φ ( 2 ) + ..... + φ ( n ) + .....
where φ ( n ) decreases as n increases is convergent or divergent according as the
∞
integral ∫ φ ( x ) dx
1
is finite or infinite.

Proof : Let S n = φ (1) + φ ( 2 ) + ..... + φ ( n )

From the above figure, it can be seen that the area under the curve y = φ ( x )
between any two ordinates lies between the set of exterior and interior
rectangles formed by the ordinates at
n = 1, 2, 3, .....n, n +1 ,......
Hence the total area under the curve lies between the sum of areas of all interior
rectangles and sum of the areas of all the exterior rectangles.
Hence

{φ (1) + φ ( 2 ) + ..... + φ ( n )} ≥ ∫ φ ( x ) dx ≥ {φ ( 2 ) + φ ( 3) + ..... + φ ( n + 1)}

n +1

∞
∴ S n ≥ ∫ φ ( x )dx ≥ S n +1 − φ (1)
1
Sequences and Series 55

Y= O ( n)

B1 c1

P1 c2
B2
P2 c3
B3
O1 P3
Bn - 1 cn cn+1
O2 Bn
O Bn+1
3
On On + 1

1 2 3 n nn +1
A1 A2 A3 An A

∞
As n → ∞, Lt S n is finite or infinite according as ∫ φ ( x )dx is finite or infinite.
1
Hence the theorem.

Solved Examples
EXAMPLE 62
∞
1
Test for convergence the series ∑ n log n
n=2
(JNTU 2003)

SOLUTION
∞ 1 ⎡ n 1 ⎤
∫ dx = Lt ⎢ ∫ = Lt [ log log x ]2 = ∞
n
dx ⎥
2 x log x n →∞
⎣ 2 x log x ⎦ n →∞

∴ By integral test, the given series is divergent.

EXAMPLE 63
∞
1
Test for convergence of the series ∑n
n =1
p
(JNTU 2003)

SOLUTION
n
∞ 1 ⎡ n 1 ⎤ ⎡ x − p +1 ⎤
∫1 x p dx = nLt
→∞ ⎢ ∫1 x p
⎣
dx ⎥ = Lt ⎢
⎦ n→∞ ⎣ − p + 1 ⎦1
⎥ ;

1
= Lt ⎡⎣ n1− p − 1⎤⎦
1 − p →∞n

1
Case (i) If p >1, this limit is finite; ∴ ∑ p is convergent.
n
56 Engineering Mathematics - I

1
Case (ii) If p < 1, the limit is in finite; ∴ ∑n p
is divergent.

1
Case (iii) If p = 1, the limit Lt log x
n →∞
n
1 = Lt ( log n ) = ∞ ; ∴
n →∞
∑n p
is divergent.

1
Hence ∑n p
is convergent if p > 1 and divergent if p ≤ 1

EXAMPLE 64
∞
n
Test the series ∑e 1
n2
for convergence.

SOLUTION
n
un = 2 = φ ( n )( say ) ;
en
φ ( n ) is +ve and decreases as n increases. So let us apply the integral test.
∞ ∞ ∞
e dt {t = x 2 , dt = 2 xdx}
1 −t
∫ φ ( x ) dx = ∫ xe dx = ∫
−x 2

1 1
21
1 1⎛ 1⎞ 1
= − e−t ∞
1 = − ⎜ 0 − ⎟ = , which is finite.
2 2⎝ e ⎠ 2e
By integral test, ∑u n is convergent.

EXAMPLE 65
∞
1 ⎛π ⎞
Apply integral test to test the convergence of the series ∑n
2
2
sin ⎜ ⎟
⎝n⎠
SOLUTION
1 ⎛π ⎞
Let φ ( n ) = sin ⎜ ⎟ ; φ ( n ) decreases as n increases and is +ve .
n 2
⎝n⎠
∞ ∞
⎛π ⎞
Let π = t
1
∫ φ ( x ) dx = ∫ x
2 2
sin ⎜ ⎟dx ;
⎝ x2
⎠ x
0
− ∫ sin tdt = cos t π0 = finite, − π 2 dx = dt ;
1 1 1 1 dx = − 1 dt
ππ π 2 π x x2 π
2

∴ By integral test, ∑u n converges x = 2 ⇒ t = π x=∞⇒t =0

2
Sequences and Series 57

EXAMPLE 66

Apply integral test and determine the convergence of the following series.
∞ ∞ ∞
3n 2n 3 1
(a) ∑1 4n2 + 1 (b) ∑1 3n4 + 2 (c) ∑ 3n + 1
1

SOLUTION
3n
(a) φ (n) = is +ve and decreases as n increases
4n 2 + 1
∞ ∞
3x ⎛ 4 x 2 + 1 = t ⇒ xdx = 1 dt ⎞
∫1 φ ( x )dx = ∫1 4 x 2 + 1 dx ⎜⎜ x = 1 ⇒ t = 5, x = ∞ ⇒8t = ∞ ⎟⎟
⎝ ⎠
∞
⎡ 3 t dt ⎤ ⎡3 ⎤
∫ φ ( x ) dx = nLt
1
→∞
⎢ ∫ ⎥ = nLt
⎣ 8 5 t ⎦ →∞ ⎣ 8
⎢ log t − log 5⎥ = ∞
⎦
∴ By integral test, ∑u n diverges.
2n 3
(b) φ ( n) = decreases as n increases and is +ve
3n 4 + 2
∞ ∞
2 x3
∫1 ( ) ∫1 3x 4 + 2 dx
φ x dx =
∞
1 dt 1
= [ log t ]5 = ∞
∞
= ∫
65 t 6
[where t = 3 x 4 + 2 ]

By integral test, ∑u n is divergent.

1
(c) φ (n) = is +ve, and decreases as n increases.
3n + 1
∞ ∞ ∞
1 1 dt 1
∫1 ( ) ∫1 3x + 1 ∫4 3 t [t = 3x + 1] = 3 log t t = ∞
φ ∞
x dx = dx =

∴ By integral test, ∑u n is divergent.

Alternating Series
A series, u1 − u2 + u3 − u4 + − + ..... + ( −1)
n −1
un + ..... , where un are all +ve, is an

alternating series.
58 Engineering Mathematics - I

1.3.7 Leibneitz Test

∞

∑ ( −1)
n −1
If in an alternating series .un , where un are all +ve,
n =1

(i) un > un +1 , ∀n, and (ii) Lt un = 0 , then the series is convergent.

n →∞

Proof :
Let u1 − u2 + u3 − u4 + ...... be an alternating series (‘ un ’are all +ve)
Let u1 > u2 > u3 > u4 ...... , Then the series may be written in each of the following
two forms :
( u1 − u2 ) + ( u3 − u4 ) + ( u5 − u6 ) + ..... .........(A)
u1 − ( u2 − u3 ) − ( u4 − u5 ) − ..... ..........(B)
(A) Shows that the sum of any number of terms is +ve and
(B) Shows that the sum of any number of terms is < u1 .
Hence the sum of the series is finite. ∴ The series is convergent.
Note : If Lt un ≠ 0 , then the series is oscillatory.
n →∞

Solved Examples
EXAMPLE 67
1 1 1
Consider the series 1 − + − .....
2 3 4
In this series, each term is numerically less than its preceding term and nth term → 0 as
n → ∞.
∴ By Leibneitz’s test, the series is convergent.
(Note the sum of the above series is Log e 2 )

EXAMPLE 68
( −1)
n −1

Test for convergence ∑ 2n − 1

(JNTU 1997)

SOLUTION
1
∑ ( −1)
n −1
The given series is an alternating series un , where un =
2n − 1
We observe that (i) un > 0, ∀n (ii) un > un +1 , ∀n (iii) Lt un = 0
n →∞

∴ By Leibneitz’s test, the given series is convergent.

Sequences and Series 59

EXAMPLE 69
1 1 1
Show that the series S = 1 − + − + ...... converges. (JNTU 2000)
3 9 27
SOLUTION
( −1)
n −1
∞
1
∑ = ∑ ( −1)
n −1
The given series is n −1
un , where un = is an alternating series
1 3 3n −1
in which 1. un > 0, ∀n 2. un > un +1 , ∀n and 3. Lt un = 0 ;
n →∞

Hence by Leibneitz’s test, it is convergent.

EXAMPLE 70
x x2 x3
Test for convergence of the series, − + − +....., 0 < x < 1
1 + x 1 + x 2 1 + x3
(JNTU 2003)
SOLUTION
( −1)
n −1
.x n
∑ = ∑ ( −1)
n −1
The given series is of the form un ,
1+ x n

xn
where un = Since 0 < x < 1, un > 0, ∀n ;
1 + xn
xn x n +1
Further, un − un +1 = −
1 + x n 1 + x n +1
x n − x n +1 x n (1 − x )
= = .
(1 + x n )(1 + xn+1 ) (1 + x n )(1 + x n+1 )
0 < x < 1 ⇒ all terms in numerator and denominator of the above expression are +ve.

∴ un > un +1 , ∀n.
0
Again, x n → 0 as x n → ∞ since 0 < x < 1; ∴ Lt un = =0
n →∞ 1+ 0
∴ By Leibneitz’s test, the given series is convergent.

EXAMPLE 71
( −1)
n −1
∞
Test for convergence ∑
n=2 n ( n + 1)( n + 2 )
(JNTU 2004)
60 Engineering Mathematics - I

SOLUTION
∑ ( −1)
n −1
The given series is an alternating series un
1
where un = ; un > 0, ∀n ;
n ( n + 1)( n + 2 )
Again, ( n + 1)( n + 2 )( n + 3) > n ( n + 1)( n + 2 )
1 1
∴ < , ∀n
( n + 1)( n + 2 )( n + 3) n ( n + 1)( n + 2 )
i.e., un +1 < un , ∀n
1
Further, Lt un = Lt =0
n →∞ n →∞
n ( n + 1)( n + 2 )
∞

∑ ( −1)
n −1
∴ By Leibnitz’s test, un is convergent
2

EXAMPLE 72
Test for the convergence of the following series,
1 2 3 4 5
− + − + − +.... (JNTU 1998, 2004)
6 11 16 21 26
SOLUTION
∞
n
∑ ( −1) = ∑ ( −1) un is an alternating series
n −1 n −1
Given series,
n =1 5n + 1
n n n +1 −1
un = > 0∀n ; − = ⇒ un < un +1 , ∀n
5n + 1 5n + 1 5n + 6 ( 5n + 1)( 5n + 6 )
n 1
Again, Lt un = Lt = ≠0
n →∞ n →∞ 5n + 1 5
Thus conditions (ii) or (iii) of Leibnitz’s test are not satisfied. The given series is not
convergent. It is oscillatory.
EXAMPLE 73
Test the nature of the following series.
( −1) ( −1) ( −1)
n −1 n −1 n −1
∞ ∞
(a) ∑ 1 n + n +1
(b) ∑ n2 + 1
(c) ∑
n =1 n +1
Sequences and Series 61

SOLUTION
1
(a) un = > 0∀n ;
n + n +1
1 1
un − un +1 = −
n + n +1 n +1 + n + 2

n+2 − n 2
= = >0
( n + n +1 )( n +1 + n + 2 ) ( n+2 + n )( n + n +1 )( n +1 + n + 2 )
∴ By Leibnitz’s test the series converges.
1 1 1
(b) un = 2 > 0, ∀n ; 2 > ⇒ un > un +1 , ∀n ;
n +1 n + 1 ( n + 1)2 + 1
Lt un = 0 ∴ By Leibnitz’s test, given series converges.
n →∞

1
(c) un = > 0, ∀n ;
n +1
1 1
n + 2 > n +1 ⇒ < ⇒ un > un +1 , ∀n
n + 2 n +1
By Leibnitz’s test, given series converges.
EXAMPLE 74
1 1 1
Test the convergence of the series − + − +...... (JNTU 2004)
5 2 5 3 5 4
SOLUTION
( −1)
n −1
∞
1
The series can be written as ∑5
n =1 n +1
; un =
5 n +1
(i) un > 0∀n
1 1
(ii) 5 n + 2 > 5 n +1 ⇒ < ⇒ un > un +1∀n
5 n + 2 5 n +1
(iii) Lt un = 0 ;
n →∞

By Leibnitz’s test, the given series converges.

EXAMPLE 75
Test for convergence the series, 1 − 1 + 1 − 1 + ..... (JNTU 1997)
2 4 6
62 Engineering Mathematics - I

SOLUTION
( −1)
n

The given series can be written as

2n
∑ (omitting 1st term)

1 1 1 1
> 0∀n ; > ⇒ un > un +1 , ∀n ; Lt =0
2n 2n 2n + 2 n →∞ 2n

( −1)
n

∴ By Leibnitz’s test, ∑ is convergent.

2n
EXAMPLE 76
1 1 1
Test for convergence the series, 1 − + − + ...... (JNTU 2004, 2007)
3! 5! 7!
SOLUTION
( −1)
n −1

General term of the series is

( 2n − 1)!
1
The series is an alternating series; > 0∀n
( 2n − 1)!
1 1 1
> ⇒ u > u , ∀n ∈ N ; Lt =0
( 2n − 1)! ( 2n − 1)! n n+1 n →∞ ( 2n − 1) !

By Leibnitz’s test, given series is convergent.

1.4 Absolute convergence

A series ∑u n is said to be absolutely convergent if the series ∑| u n | is convergent
Ex. Consider the series
1 1 1
∑u n = 1−
+ − + − + ......
23 33 43
∞
1 1 1 1
∑ n u = 1 + 3
+
2 3 43
+ 3
+ ...... = ∑
1 n
3

By p − series test, ∑ un is convergent ( p = 3 > 1 )

Hence ∑u n is absolutely convergent.
Note: 1. If ∑u n is a series of +ve terms, then ∑u n = ∑| u n |.
For such a series, there is no difference between convergence and absolute
convergence. Thus a series of +ve terms is convergent as well as absolutely
convergent.
Sequences and Series 63

2. An absolutely convergent series is convergent. But the converse need not be

true.
∞
1 1 1 1
Consider ∑ (−1)
1
n −1
. = 1 − + − + ......
n 2 3 4
This series is convergent (1.7.3)
1 1 1
But ∑ (−1) . 1 n = 1 + 2 + 3 + 4 + ....... is divergent (p-series test).
n −1

Thus ∑ u is convergent need not imply that ∑ | u | is convergent (i.e.,

n n

∑ u is not absolutely convergent).

n

1.5 Conditional Convergence

If the series ∑u n is divergent and ∑u n is convergent, then ∑u n is said to be
conditionally convergent.
Ex. Consider the Series
1 1 1
1 − + − ....... ∑ un is convergent by Leibnitz’s test. (Ex.1.7.3)
2 3 4
1 1 1
But ∑ un = 1 + + + ..... is divergent by p − series test.
2 3 4
∴ ∑ un is conditionally convergent.

1.6 Power Series and Interval of Convergence

A series, a0 + a1 x + a2 x 2 + ..... + an x n + ..... where ‘ an ’ are all constants is a power
series in x .
It may converge for some values of x .
un +1 a an +1
Lt = Lt n +1 .x (1st term is omitted.) = kx (say) where Lt =k
n →∞ u n →∞ a n →∞ a
n n n

Then, by ratio test, the series converges when kx < 1 .

⎛ −1 1 ⎞
i.e., it converges ∀x ∈ ⎜ , ⎟ (k ≠ 0)
⎝ k k⎠
64 Engineering Mathematics - I

⎛ −1 1 ⎞
The interval ⎜ , ⎟ is known as the interval of convergence of the given power
⎝ k k⎠
series.

Solved Examples
EXAMPLE 77
∞
xn
Find the interval of convergence of the series ∑
n =1 n
3

SOLUTION
xn x n +1
un = ; u n +1 =
n3 (n + 1)3
3
⎛ ⎞
⎛ un +1 ⎞ ⎛ n ⎞
3
⎜ 1 ⎟
Lt ⎜ ⎟ = Lt ⎜ ⎟ .x = Lt ⎜ ⎟ .x = x
n →∞
⎝ un ⎠ n →∞ ⎝ n + 1 ⎠ n →∞
⎜ 1+ 1 ⎟
⎝ n⎠
By ratio test, the given series converges when x < 1, i.e., x ∈ (−1,1)
When x = 1, ∑ u = ∑ 1n
n 3 , which, is convergent by p series test.
∴ ∑u n is convergent when x = 1
Hence, the interval of convergence of the given series is (–1, 1)
EXAMPLE 78
Test for the convergence of the following series.
(a) 1− 1 + 1 − 1 + .......... (JNTU 1996)
2 3 4
(b) 1 + 1 2 − 1 2 − 1 2 + 1 2 + 1 2 − 1 2 − 1 2 + ....... (JNTU 1998)
2 3 4 5 6 7 8
1 − x + x − x + ..........
2 4 6
(c) (JNTU 2004)
2 4 6
∞
1
(d) ∑0 (−1)n (n + 1) x n , with x < 2 (JNTU 2004)

SOLUTION
(a) The series is of the form ∑ (−1) n −1
un where un = 1
n
Sequences and Series 65

It is an alternating series where (i) un > 0∀n (ii) un > un +1∀n and
(iii) Lt un = 0;∴ By Leibnitz test, the series is convergent.
n →∞

Again the series 1+ 1 + 1 + 1 + ...... is divergent, by

2 3 4
p − series test.
Hence the given series is conditionally convergent.
∞
(b) ∑ un = ∑ 1 p =1
n2
which is convergent by p − series test .

∴ The given series is absolutely convergent.

∴ It is convergent.
(c) The given series is
x 2n−2 x 2n−2
∑ (−1) . (2n − 2)!. =∑ (-1) un ; ∴ un = (2n − 2)!
n −1 n −1

x2n un +1 1 u
un +1 = ; = . x 2 ; Lt n +1 = 0 < 1
2n ! un (2n − 1)(2n) n →∞ un
By ratio test, the series ∑u n converges ∀x; i.e., ∑u n is absolutely
convergent ∀x;
(d) Here, un = ( n + 1) x n ;| un +1 |= ( n + 2) x n +1 (neglect 1st term)

un +1 (n + 2) (1 + 2 )
Lt = Lt x = Lt n x = x < 1 (∵ x < 1 )
n →∞ ( n + 1) 2
n →∞ u n →∞
(1 + )
1
n
n
∴ ∑ un is convergent ∀x , i.e., given series is absolutely convergent and
hence convergent.
EXAMPLE 79
Show that the series 1 + x + x +x
2 2
+ ..... converges absolutely ∀x
2 3
SOLUTION

un +1 x x n −1 xn
Lt = = 0 < 1 when x ≠ 0 [since un = ; un +1 = ]
n →∞ u
n n (n − 1)! n!

∴ By ratio test, ∑u n is convergent ∀x ≠ 0 .

66 Engineering Mathematics - I

When x = 0 , the series is (1 + 0 + 0 + ......) and is convergent

∴ ∑u n converges ⇒ ∑u n is absolutely convergent ∀x .

EXAMPLE 80
1 1 1
Show that the series, 1 − + − + ........ is absolutely convergent.
3 32 34
SOLUTION
∞
1
∑ un = ∑ 1 n =1
3n −1 , which is a geometric series with common ratio
3
<1

∴ It is convergent. Hence given series is absolutely convergent.

EXAMPLE 81
Test for convergence, absolute convergence and conditional convergence of the series,
1 1 1
1 − + − + .......... (JNTU 2003)
5 9 13
SOLUTION
1
The given alternating series is of the form ∑ (−1) n −1
un , where, un =
4n − 3
.

1 1
Hence, un > 0∀n ∈ N ; un +1 = =
4 ( n + 1) − 3 4n + 1
1 1
un − un +1 = −
4n − 3 4n + 1
4n + 1 − 4n + 3 4
= = > 0, ∀n ∈ N
(4n − 3)(4n + 1) (4n − 3)(4n + 1)
1
i.e., un > un +1 , ∀n ∈ N Lt un = Lt = 0;
n →∞ n →∞ 4n − 3
All conditions of Leibnitz’s test are satisfied.
Hence ∑ (−1) n −1
un is convergent.

1 1 un n 1
un = ; Take vn = ; Lt = Lt = ≠ 0 and finite.
4n − 3 n n→∞ vn n→∞ n(4 − 3 ) 4
n
∴ By comparison test, ∑u n and ∑v n behave alike.
Sequences and Series 67

But by p − series test, ∑v n is divergent (since p = 1) .

∑u n is divergent and ∴ The given series is conditionally convergent.

EXAMPLE 82
∞
1
Test the series ∑ (−1)
n =1
n −1
.
3 n
, for absolute / conditional convergence .

SOLUTION
∑ ( −1)
n −1
The given series is an alternating series of the form un .

Here
1
(i) un = , ∀n ∈ N
3 n
(ii) 3 ( n + 1) > 3n ⇒ 3 n + 1 > 3 n , ∀n .
1 1
∴ < , i.e., un +1 < un , ∀n ∈ N
3 n +1 3 n
1
And Lt un = Lt =0
n →∞ n →∞ 3 n

∴ By Leibnitz’s test, the given series is convergent.

1 1
But ∑ (−1) n −1. =∑ is divergent by p − series test (since
3 n 3 n
1
p = <1 )
2
∴ The given series is conditionally convergent.

EXAMPLE 83

Test the following series for absolute / conditional convergence.

∞
sin(nα ) ∞
n2
(a) ∑ (−1)
n =1
n −1
,
n2
(b) ∑ (−1)
n =1
n −1
.
n3 + 1
∞
(−1) n
nπ n
(c) ∑
n = 0 (2n)!
(d) ∑ (−1) . e3n+1
n −1
68 Engineering Mathematics - I

SOLUTION
sin nα 1
(a) un = < 2 ⎡⎣since sin nα < 1⎦⎤ considering vn = 1 2 and using
n 2
n n
comparison and p - series tests, we get that ∑ un is convergent ∑ un is
absolutely convergent .
(b) By Leibnitz’s test, the series converges.
1 n2
Taking vn =
n
, by comparison and p - series tests , ∑ n3 + 1 , is seen to
be divergent.
Hence given series is conditionally convergent.
1 un +1
(c) Take un = ; Lt
2n ! n→∞ un
= 0 < 1 ; By ratio test, ∑u n is convergent;

Hence given series is absolutely convergent.

nπ n
(d) un = ; By root test, is convergent, ∴ given series is absolutely
e3n +1
convergent.
[In problems (a) to (d) above, hints only are given. Students are advised to
do the complete problem themselves]
EXAMPLE 84
Find the interval of convergence of the following series.
∞ ∞
n( x + 2) n
∑ (−1)n−1 x ∑ (−1)n−1
n
(a) (b) (c) log(1 + x)
n =1
n3 n =1 3n
SOLUTION
xn x n +1
(a) Let the given series be ∑ un ; Then un = n3
; un +1 =
(n + 1)3
3
⎛ ⎞
un +1 ⎛ n ⎞
3
⎜ 1 ⎟
Lt = Lt ⎜ ⎟ .x = Lt ⎜ ⎟ .x = x
un n →∞ n + 1
n →∞
⎝ ⎠ n →∞ 1
⎜ 1+ ⎟
⎝ n⎠
∴ By ratio test, ∑u n is convergent if x < 1

i.e., ∑u n is absolutely convergent if x < 1 ;

Sequences and Series 69

∴ ∑u n is convergent if x < 1

1 1 1
If x = 1, the given series becomes 1 − + − + ........
23 33 43
1
which is convergent, since ∑n 3
is convergent.

1 1
Similarly, if x = –1, the series becomes ∑− n 3
= −∑
n3
which is also

convergent.
Hence the interval of convergence of ∑u n is ( −1 ≤ x ≤ 1)
(b) Proceeding as in (a),
un +1 x+2
Lt =
n →∞ un 3

∴ ∑u n is convergent if x + 2 < 3 , i.e., if −3 < x + 2 < 3 , i.e., if

−5 < x < 1 .

∑ u = ∑ ( −1)
2 n −1
If x = –5, n .n , and is divergent ( in both these cases

If x = 1, ∑ u = ∑ ( −1)
n −1
n .n , and is divergent Lt un ≠ 0 )
n →∞

Hence the interval of convergence of the series is (–5 < x < 1)

x 2 x3 x 4
(c) log (1 + x ) = x − + − + .......
2 3 4
∞
xn
∑ ( −1) = ∑ un
n −1
= (say)
n =1 n
xn x n +1
un = ; un +1 =
n n +1
70 Engineering Mathematics - I

un +1 1
Lt = Lt | x |= x
n →∞ un n →∞ ⎛ 1⎞
⎜1 + ⎟
⎝ n⎠
By ratio test, ∑u n is convergent when x < 1

i.e., ∑u n is absolutely convergent and hence convergent when –1 < x < 1.

1 1 1 1
∑ u = ∑ ( −1)
n −1
When x = –1, n . = 1 − + − + ....... ,
n 2 3 4

which is convergent by Leibnitz’s test. (give the proof )

⎛ 1 1 1 ⎞
When x = 1, ∑u n = − ⎜1 +
⎝
+ + + ...... ⎟ which is divergent,
2 3 4 ⎠
1
since ∑n is divergent by p –series test ( prove ) .

Hence ∑u n is convergent when −1 < x ≤ 1 .

Interval of convergence is ( −1 < x ≤ 1 ).

Exercise 1.5
1. Use integral test and determine the convergence or divergence of the following
series:
1
1 ∑n 2
...................................... [Ans : convergent]
∞
1
2. ∑n ....................................... [Ans : convergent]
( log n )
2
2

2. Test for convergence of the following series:

1 1 1
1. 1 − + − + .... .................................. [Ans : convergent]
2 4 6
( −1)
n −1
∞
2. ∑1 ( 2n − 1)( 2n ) ................................... [Ans : convergent]
Sequences and Series 71

∑ ( −1)
n −1 −5
3. n 2
................................... [Ans : convergent]
1

3. Classify the following series into absolutely convergent and conditionally

convergent series :
( −1)
n

1. ∑ n3
....................................... [Ans : abs.cgt]

sin n
2. ∑ 3
.......................................... [Ans : abs.cgt]
n 2

( −1)
n

3. ∑ n log n 2 ......................................... [Ans : abs.cgt]

( )
4. Find the interval of convergence of the following series :
2n x n
1. ∑ n ............................................ [Ans : −∞ < x < ∞ ]

xn
2. ∑ n2 ............................................ [Ans : −1 ≤ x ≤ 1 ]

x2 x3 x 4
3. x − + − + ...... .................................. [Ans : −1 < x ≤ 1 ]
2 3 4
1 1 1
5. (a) Show that 1 − 2 + 2 − 2 + .... is absolutely convergent.
2 3 4
1 1 1
(b) Show that 1 − + − + .... is conditionally convergent.
2 3 4

Summary
∞
1. The geometric series ∑x
n =1
n −1
converges if | x |< 1 , diverges if x ≥ 1 , and oscilates

when x ≤ −1
2. If ∑
un is convergent, Lt un = 0 [The convergent need not be necessary ]
n →∞
∞
1
3. p – series test :- ∑n
n =1
p
is convergent if p > 1 and divergent if p ≤ 1
72 Engineering Mathematics - I

4. Comparison test :- The series ∑u n and ∑v

n are both convergent or both
un
divergent if Lt is finite and non – zero.
n →∞ v
n

5. D’Alembert’sRatio test :- ∑u n converges or diverges according as

un +1
Lt < 1 or > 1
n →∞ un
⎛ un ⎞
⎜ Alternately, if nLt > 1 or < 1⎟ . If the limit = 1, the test fails
→∞ u
⎝ n +1 ⎠
6. Raabe’s test : ∑ un converges or diverges according as
⎡ ⎧u ⎫⎤
Lt ⎢ n ⎨ n − 1⎬⎥ > 1 or < 1 .
⎣ ⎩ un +1 ⎭⎦
n →∞

7. Cauchy’s root test: ∑ un converges or diverges according as

⎛ 1⎞
Lt ⎜ un n ⎟ < 1 or > 1 (If limit = 1, the test fails.)
n →∞
⎝ ⎠
8. Integral test : A series ∑ φ ( n ) of +ve terms where φ ( n ) decreases as n increases
∞
is convergent or divergent according as the integral φ ( x ) dx is finite or infinite.
∫
1
∞

∑ ( −1)
n −1
9. Alternating series – Leibnitz’s test: An alternating series un convergent
n =1

if (i) u n > u n +1 , ∀n and (ii) Lt un = 0

n →∞

10. Absolute / conditional convergence :

(a) ∑
un is absolutely convergent if ∑ | u | is convergent.
n

(b) ∑u n is conditionally convergent if ∑ u is convergent and ∑ | u

n n | is

divergent.
(c) An absolutely convergent series is convergent, but converse need not be true .
i.e., a convergent series need not be convergent.
Sequences and Series 73

Miscellaneous Exercise - 1.6

1. Examine the convergence of the following series:
1 1 1
1. + + + ...... ................................... ... [cgt.]
1.2 3.4 5.6
12 22 32
2. + + + ...... .............................. [dgt.]
13 + 1 23 + 1 33 + 1
2 2 2 23
3. + + + ....... .............................. [dgt.]
1 2 3
1 2 3
4. + + + ..... ................................ [cgt.]
1 + 2 1 + 2 1 + 23
2

x x2 x3
5. + + + ..... ( x > 0 ) ........ [cgt. if x ≤ 1 dgt. if x > 1]
1 + x 1 + x 2 1 + x3
3x 2 4 x3
6. 2x + + + ..... ( x > 0 ) .......................... [cgt. if x ≤ 1 dgt. if x > 1]
8 27
1 1.3 1.3.5
7. 1+ + + + .... ................................. [dgt.]
2 2.4 2.4.6
32 32.52 32.52.7 2
8. + + + ...... ............................... [cgt]
62 62.82 62.82.102
3.4 4.5 5.6
9. + + + .... ................................ [dig.]
1.2 2.3 3.4
( 1) ( 2) ( 3)
2 2 3

10. .x + x 2
+ x3 + .... ( x > 0 ) ............. [cgt. if x <4, dgt. if x ≥ 4 ]
2 4 6

x x 2 x3
11. 1+ + + + ..... ( x > 0 ) ......................... [cgt.if x ≤ 1 , dgt. if x > 1]
22 32 42
2 3
3x ⎛ 4 ⎞ 2 ⎛ 5 ⎞
12. + ⎜ ⎟ x + ⎜ ⎟ + x 3 + ..... ( x > 0 ) ............. [cgt if x < 1, dgt. if x ≥ 1 ]
4 ⎝5⎠ ⎝6⎠
n2
⎛ 1⎞
13. ∑ ⎜⎝1 + n ⎟⎠ .......................... [dgt.]
74 Engineering Mathematics - I

23 n
14. ∑ 32n ................................................ [cgt.]

an
15. ∑ 1 + n2 , a < 1 ................................................ [cgt.]

1 1 1
16. 1− + − + − + −..... ....................... [Abs. cgt.]
2.2 3.3 4.4

2. Examine for absolute and conditional convergence of the following series:

33n
∑ ( −1)
n
1. .................................... [Abs. cgt.]
32 n
( −1)
n
.n
2. ∑ 2 n
...................................... [Abs. cgt.]

1 1 1 1
3. 1− + − + ........................... [Cond. cgt]
2 3 4 5
n ( n + 1)
2

4. ∑ ( −1) n3 .............................. [Cond.cgt.]

3. Determine the interval of convergence of the following series :

x 2 x3 x 4
1. x+ + + + ..... ............................... [ −1 ≤ x < 1 ]
2 3 4
( x + 1)
n

2. ∑ n.2n
................................ [ −3 < x < 1 ]

Solved University Questions (J.N.T.U)

1. Test the convergence of the series:
1 2 3
+ + + .........
1.2.3 2.3.4 3.4.5
Solution
Let un be the nth term of the series;
n 1
Then, un = =
n ( n + 1)( n + 2 ) ( n + 1)( n + 2 )
Sequences and Series 75

1 un n2
Let vn = 2 ; then, Lt = Lt
n n →∞ v n →∞ ( n + 1)( n + 2 )
n

1
= Lt = 1,
n →∞ ⎛ 1 ⎞⎛ 2 ⎞
⎜1 + ⎟ ⎜1 + ⎟
⎝ n ⎠⎝ n ⎠
Which is non-zero and finite.
∴ By comparison test, both ∑u n and ∑v n converge or diverge together.

But ∑v n is convergent by p-series test (p > 1) ∴ ∑u n is convergent.

2. Show that every convergent sequence is bounded

Solution
Let an be a sequence which converges to a limit ‘l ‘say.

Lt an = l ⇒ given any +ve number ∈ , however small ,

n →∞

we can always find an integer ‘ m ‘ , ∋ , an − l <∈ , ∀n ≥ m

Taking ∈ = 1, we have, an − l < 1 ;

i.e., ( l − 1) < an < ( l + 1) , ∀n ≥ m

{ } {
Let λ = min a1 , a2 ,........am −1 , ( l − 1) , and μ = max a1 , a2 ,........am −1 , ( l + 1) }
Then obviously, λ ≤ an ≤ μ , ∀n ∈ N ;

Hence an is bounded.
∞
3. Show that the geometric series ∑q
m=0
m
= 1 + q + q 2 + ....... converges to the sum

1
when q < 1 and diverges when q ≥ 1
1− q
(JNTU 2001)
Solution
See theorm 1.2.3 (replace ‘x ‘ by ‘q’ ) .
76 Engineering Mathematics - I

4. Define the convergence of a series. Explain the absolute convergence and

conditional convergence of a series. Test the convergence of series

(JNTU 2000)
− n2
⎡ 1 ⎤
∑ ⎢⎣1 + ⎥
n⎦
Solution
For theory part , refer 1.2.1, 1.2.2, 1.8, 1.8.1, 1.9.1 and 1.9.2

( )
− n2 −n
⎛ 1 ⎞ 1 ⎛ 1 ⎞
Problem : Let un = ⎜1 + ⎟ ; Lt u n
n
= Lt ⎜1 + ⎟
⎝ n⎠ n →∞ n →∞
⎝ n⎠
1 1
= Lt n
= 2 <1
n →∞
⎡ 1 ⎤ e
⎢1 + n ⎥
⎣ ⎦
By Cauchy’s root test, ∑ un is convergent.
1 1.3 2 1.3.5 3
5. Test the convergence of the series, 1 + x+ x + x + .....
2 2.4 2.4.6
(JNTU 2001)
Given that x > 0.
Solution
Omitting the first term of the series, we have,
1.3.5. ( 2n − 1) n 1.3.5..... ( 2n + 1) n +1
un = x ; un +1 = .x ;
2.4.6....2n 2.4.6..... ( 2n + 2 )
un +1 ⎛ 2n + 1 ⎞
Lt = Lt ⎜ ⎟ .x = x
n →∞ un n →∞
⎝ 2n + 2 ⎠
By ratio test, ∑u n is convergent when x < 1, and divergent when x > 1
The ratio test fails when x = 1
un 2n + 2 1
When x = 1, −1 = −1 =
un +1 2n + 1 2n + 1
⎡ ⎛ u ⎞⎤ ⎛ n ⎞ 1
Lt ⎢ n ⎜ n − 1⎟ ⎥ = Lt ⎜ ⎟ = <1 ;
u n →∞ 2n + 1
n →∞
⎣ ⎝ n +1 ⎠ ⎦ ⎝ ⎠ 2
∴ By Raabe’s test , ∑ un diverges .
∴ The given series converges when x < 1 and diverges when x ≥ 1 .
Sequences and Series 77

2 3
1 ⎛2⎞ ⎛3⎞ ⎛4⎞
6. Test the convergence of the series, + ⎜ ⎟ x + ⎜ ⎟ x 2 + ⎜ ⎟ x 3 + .........x > 0
2 ⎝3⎠ ⎝4⎠ ⎝5⎠
(JNTU 2002)
Solution
Neglecting the 1st term,
n
⎡⎛ n + 1 ⎞ ⎤
u n = ⎢⎜ ⎟ x⎥ ;
⎣⎝ n + 2 ⎠ ⎦
⎛ 1+ 1 ⎞
1 ⎛ n +1 ⎞ ⎜ n⎟x
un n = ⎜ ⎟ x =
⎝ n + 2 ⎠ ⎜ 1+ 2 ⎟
⎝ n⎠
∑u
1
Lt un n = x ; By Cauchy’s root test, n is cgt. when x < 1 and dgt. when
n →∞
x > 1; when x = 1 , the test fails .

(1+ 1 )
n

n e 1
When x = 1, un = ; Lt un = = ≠0
(1 + 2 n )
n n →∞ e2 e

∴ ∑u n is divergent.
∴ ∑u n is cgt. when x < 1 and dgt. when x ≥ 1 .

7. Test the series whose nth term is ( 3n − 1) 2n for convergence. (JNTU 2003)
Solution

un =
( 3n − 1) ; un +1 =
{3 ( n + 1) − 1} ;
2n 2n +1
un +1
=
( 3n + 2 ) Lt
un +1 1
= <1 ;
un 2 ( 3n − 1) n →∞ u
n 2
∴ By ratio test, ∑u n is convergent.
∞
1
8. Show by Cauchy’s integral test that the series ∑n converges if p > 1
( log n )
p
n=2

and diverges if 0 < p ≤ 1 (JNTU 2003)

Solution
1
Let φ ( x ) = ; x ≥ 2 ; Then φ ( x ) decreases as x increases in [ 2, ∞ ]
x ( log x )
p
78 Engineering Mathematics - I

∞ ∞ ∞
dx du u1− p
∫ φ ( x )dx = ∫ x ( log x ) = ∫ p = ∞
;
p
1− p
log 2
2 2 log 2
u
1
[Taking log x = u, dx = du x = 2 ⇒ u = log 2 and x = ∞ ⇒ u = ∞ ]
x
Case (i) : p > 1 ⇒ 1 − p < 0 ⇒ Integral is finite , and
Case (ii) : 0 < p ≤ 1 ⇒ Integral is infinite.
Hence, by integral test, the given series converges if p > 1 and diverges when
0 < p ≤1.
3
n 2
⎛ 1 ⎞
9. Test the convergence of the series ∑ ⎜⎜⎝1 + ⎟⎟
n⎠
Solution
1
⎧ n 2⎫
3 n
1 ⎪⎛ 1 ⎞ ⎪ 1
u n n ⎨⎜⎜1 + ⎟⎟ ⎬ = ;
n
⎪ ⎝ n ⎠ ⎪ ⎛ 1 ⎞
⎩ ⎭ ⎜⎜1 + ⎟⎟
⎝ n⎠
1 1
Lt un n = <1 [2<e<3].
n →∞ e
By Cauchy’s root test, ∑u n is convergent.

( −1) .x n
n
∞
10. Test the convergence of the series, ∑ ,0< x <1
n = 2 n ( n − 1)

Solution
xn
∑ ( −1)
n
The given series is of the form un , where un = .
n ( n − 1)
This is an alternating series in which (i) un > 0 and un > un +1∀n ∈ N .
Further Lt un = 0 . Hence, by Leibnitz test, the series is convergent.
n →∞

1 x2 x4 x6
11. Discuss the convergence of the series, + + + + ..........
2 1 3 2 4 3 5 4
(JNTU 1995, 2002, 2003, 2008)
Solution
x2n
nth term of the series = un = (omitting 1st term)
( n + 2) n + 1
Sequences and Series 79

x 2n+2 u n + 2 n +1 2
un +1 = ; n +1 = .x
( n + 3 ) n + 2 un ( n + 3)
⎡ 1+ 2 . 1+ 1 ⎤
un +1 ⎢ n n 2⎥
Lt = Lt .x = x 2 ;
n →∞ u
n
n →∞ ⎢
⎢⎣
1+ 3
n
⎥
(
⎥⎦ )
∴ By ratio test, ∑ un converges if x 2 < 1 , i.e., if x < 1 , and diverges if
x 2 > 1 , i.e., if x > 1 ;
1 1
When x 2 = 1 , un = ; taking vn = ,
( n + 2) n +1 n
3
2

3
un n 2
Lt = Lt 3 =1
n →∞ v
n
n →∞
n 2 1+ 2
n
1+ 1
n ( )
∴ By comparison test, ∑ un and ∑ vn both converge or diverge together;
∑v
But n is convergent by p-series test.
∴ ∑u n is convergent if x ≤ 1 and divergent if x > 1 .

∞
x2n
12. Test the convergence of the series ∑
n =1 ( n + 1) n
(JNTU 2006, 2007)

Solution
x2n x2n+2
un = ; un +1 =
( n + 1) n ( n + 2) n + 1
un +1 n n +1 2 1+ 1
= .x = n .x 2 ; Lt un +1 = x 2 ;
un n+2 1+ 2
n
n →∞ u
(n )
∴ By ratio test, ∑ un converges when x < 1 and diverges for x > 1 .
1 1
When x = 1 , un = taking vn =
( )
3 3
and applying the comparision
n 2
1+ 1 n 2
n
test, we observe that ∑u n is convergent .
Hence ∑u n converges when x ≤ 1 and diverges when x > 1 .
80 Engineering Mathematics - I

x 2 x3 x 4
13. Find the interval of convergence of the series, + + + .........∞
2 3 4
(JNTU 2006, 2007)
Solution
x n +1 xn+2
For the given series, un = ; un +1 =
n +1 n+2
u ⎛ 1+ 1 ⎞
Lt n +1 = Lt ⎜ n⎟x = x
n →∞ ⎜
n →∞ u
1+ 2 ⎟
n
⎝ n⎠
By ratio test, ∑ un converges when x < 1 i.e., −1 < x < 1
1
When x = 1, un =
n +1
1 u 1
Taking un = ; n =
n vn 1 + 1
n
un
and, Lt = 1 ≠ 0 and finite.
n →∞ v
n

∴ Both ∑u n ∑ v converge or diverge together.

and n

But ∑ v diverges ∴ ∑ u also diverges when x = 1.

n n

When x = –1, the given series is

1 1 1 1
− + − + which is alternating series with
2 3 4 5
un > un +1∀n and un → 0 as n → ∞
∴ By Leibneitz’s test ∑u n converges when x = –1

∴ Interval of convergence is [(–1, 1) i.e., −1 ≤ x < 1

∞
1.3.5.... ( 2n + 1)
14. Test the convergence of the series ∑ (JNTU 2006)
n =1 2.5.8.... ( 3n + 2 )

Solution
1.3.5.... ( 2n + 1) 1.3.5.... ( 2n + 3)
un = ; un +1 =
2.5.8.... ( 3n + 2 ) 2.5.8.... ( 3n + 5 )
Sequences and Series 81

un +1 2n + 3
= Lt
un +1
= Lt ⎢
⎡2+ 3 ⎤
n ⎥ = 2 <1 ( )
( )
;
n →∞ ⎢
un 3n + 5 n →∞ u
n 3+ 5 ⎥ 3
⎣ n ⎦
∴ By ratio test, ∑u n is convergent.
∞
(– 1)n
15. Prove that the series ∑
n =2 n(log n )
2
converges absolutely. (JNTU 2006)

Solution
∞ ∞
1 dx dt
un =
n ( log n )
3
; ∫ x ( log x )
2
3
= ∫
log 2
t2
−1 ∞ 1
(where t = log x ) = log 2 = , which is finite.
t log 2
∴ By integral test ∑u n is convergent.
∴ ∑u n converges absolutely.
Note: This problem can also be done by Leibnitz test. The reader is advised to try
that method also.
( 2n + 1) x n , x > 0
16. Test the convergence of the series ∑ n3 + 1
(JNTU 2006)

Solution
2n + 1 n
nth term of the given series , un = x ;
n3 + 1
⎡ 2 ( n + 1) + 1 ⎤ n +1 2n + 3
un +1 = ⎢ ⎥x = x n +1
⎢⎣ ( n + 1) + 1 ⎥⎦ ( n + 1) + 1
3 3

Lt
un +1
= Lt
( 2 n + 3 ) .x
n+1

× n
( n3 + 1)
n →∞ u
n
n →∞
{
( n + 1) + 1 x ( 2n + 1)
3
}
) ( )
⎡ ⎤
Lt ⎢
⎢ 2n 1 + 3 (
2n
.n3 1 + 1 3
n
⎥
⎥ x = .x
n →∞

⎢⎣ ⎨⎩ ( n ⎬
n3 ⎭
)
⎢ n3 ⎧ 1 + 1 3 + 1 ⎫ .2n 1 + 1 ⎥
2n ⎥⎦ ( )
By ratio test, ∑ un converges if x < 1 and diverges if x > 1. If x = 1 the test fails.
82 Engineering Mathematics - I

2n + 1 1
When x = 1, un = ; Taking vn = 2 ;
n +1
3
n
un 2n + 1 2
Lt = Lt 3 × n = 2 ≠ 0 and finite
n →∞ v n →∞ n + 1
n

∴ ∑ u and ∑ v converge or diverge together .

n n

But ∑ v converges ∴ ∑ u also converges.

n n

Thus, ∑ u converges when x ≤ 1 and diverges when x > 1.

n

( −1) ( log n ) , for absolute/conditional convergence

n
∞
17. Test the series ∑
n =1 n2
(JNTU 2006)
Solution

( −1) ( log n ) ( log n )

n

un = ; un = ;
n 2
n2
∞ ∞
log x
∫2 x 2 dx = log∫ 2 te dt [taking log x = t , x = e , 1 x dx = log t ]
−t t

1
= −te− t + e− t ∞
log 2 = 0 − [1 − log 2].e− log 2 = ( log 2 − 1) ,
2
which is finite.
∴ By integral test ∑u n is convergent ⇒ ∑u n converges absolutely.
(Note that ∑u n is cgt. by Leibnitz’s test).
1
18. Test the convergence of the series ∑ (JNTU 2006)
( log log n )
n

Solution
1
Given that un = ;
( log log n )
n

1 ⎡ 1 ⎤
Lt un n = Lt ⎢ ⎥ = 0 <1
n →∞ log log n
n →∞
⎣ ⎦
By Cauchy’s root test, ∑ un is convergent.
Sequences and Series 83

19. Find the interval of convergence of the series,

1 x3 1 3 x5 1.3.5 x 7
x+ . + . . + . + ...... (JNTU 2007)
2 3 2 4 5 2.4.6 7
Solution
1.3.5..... ( 2n − 1) x 2 n +1
Term of the series, un = . (neglecting 1st term)
2.4.6.....2n ( 2n + 1)
1.3.5..... ( 2n − 1)( 2n + 1) x 2 n +3
un +1 = . ;
2.4.6.....2n ( 2n + 2 ) ( 2n + 3)

Lt
un +1
= Lt ⎢
⎡ ( 2n + 1)
2
⎤
.x 2 ⎥
Lt ⎢
⎡ n2 4 + 4 + 1
⎢ n (n 2
⎤
2⎥
)
.x ⎥ = x 2
n →∞ u
n
n →∞ ( 2n + 2 )( 2n + 3 )
⎢⎣ ⎥⎦
n →∞
(
⎢⎣ n 4 + n + n 2
2 10 6
⎥⎦ )
By ratio test, ∑ un converges when x 2 < 1 , i.e., x < 1 ⇒ −1 < x < 1
When x 2 = 1 , the test fails;
un ⎛ 4n 2 + 10n + 6 ⎞ 8n + 5
Then −1 = ⎜ − 1⎟ = 2
un +1 ⎝ 4n + 2n + 1 ⎠ 4n + 2n + 1
2

⎡ ⎛ un
Lt ⎢ n ⎜
⎞⎤
− 1⎟ ⎥ = Lt
n2 8 + 5
n (= 2 >1
)
n →∞ u
⎣ ⎝ n +1 ⎠ ⎦
n →∞
n 4+
2 2
(
n
+ 21
n )
∴ By Raabe’s test, ∑ un converges when x = 1 , i.e., x = ±1 .
2

∴ Interval of convergence of ∑u n is [–1 < x < 1]

1 2 3 n
20. Test the series + + + ... + 2 , for convergence or divergence.
2 3 8 n –1
[JNTU 2006]
Solution
n 1
un = ; Let νn = 3/2
n +12
n
un n .n 3/2
n 2
1
= = 2 =
νn n +1
2
n +1 1+ 1
n2
84 Engineering Mathematics - I

⎛ ⎞
un ⎜ 1 ⎟
Lt = Lt ⎜ ⎟ = 1 which is non-zero finite number
n →∞ ν
⎜ 1+ 12 ⎟
n →∞
n
⎝ n ⎠
∴ By comparison test, Σun and Σνn behave alike.
⎛ 3 ⎞
But Σνn is convergent by p-series test ⎜∵ p = > 1⎟
⎝ 2 ⎠
Hence Σun is convergent
2 −1 3 −1 4 −1
21. Test the convergence of the series, + 2 + 2 + .......
3 −1 4 −1 5 −1
2

[JNTU 2007]
Solution
n +1 − 1 1
un = ; Let νn =
(n + 2) 2 – 1 n3/2
[∵ Highest degree of n in Dr – Nr = 2 – 1/2 = 3/2]
1 1
un
=
(
n3/2 . n + 1 − 1 ) =
n2 1
n n
νn (n + 2) 2 – 1 4 3
n2 1
n n2
un
Lt = 1 ⇒ Σun and Σνn both converge or diverge together (by comparison
n →∞ νn
⎛ 3 ⎞
test). But Σνn converges by p-series test ⎜∵ p = > 1⎟ . Hence Σun is
⎝ 2 ⎠
convergent.

22. Test the convergence of Σ n3 + 1 − n 3 [JNTU 2008]

Solution

n3 1 n3 n3 1 n3
un can be written as, un
n3 1 n3

1 1
i.e. un = ; Let νn =
⎡ 1 ⎤ n3/ 2
n3 ⎢ 1+ 3 +1 ⎥
⎣ n ⎦
Sequences and Series 85

un 1 u 1
Then, = ⇒ Lt n = ≠ 0.
νn 1 n →∞ ν 2
n
1+ 3 +1
n
∴ By comparison test, Σun and Σνn have same property
⎛ 3 ⎞
Σνn is convergent by p-series test ⎜∵ p = > 1⎟ . Hence Σun is convergent.
⎝ 2 ⎠
x x 2 x3
23. Test for the convergence of the series, 1 + + + + ....x > 0
2 5 10
[JNTU 1998, 1985, 2002, 2002]
Solution
Neglecting the first term, we observe that the nth term of the series,
xn x n +1
un = ; un+1 = 2 , so that,
n +12
n + 2n + 2
un +1 ⎛ n 2 +1 ⎞
Lt = Lt ⎜ 2 ⎟x = x
n →∞ un n →∞ ⎝ n + 2n + 2 ⎠

∴ By ratio test, Σun converges when x < 1 and diverges when x > 1 and the test
fails when x = 1
1 1
∴ when x = 1, un = ; Taking Σνn = 2 and using comparison test, we can
n +12
n
show that Σun is convergent. [This part of proof is left to the reader as an
exercise].
∴ Σun converges for x < 1 and diverges for x > 1.
n
24. Test the convergence of Σ . x n (x > 0) [JNTU 2003]
n +12

Solution
un +1 n +1 n2 + 1
Lt = Lt . .x
n →∞ un n →∞ n 2 + 2n + 2 n
1
1+ .x
= Lt 1+
1
. n2 = 1. x = x
n →∞ n 2 2
1+ + 2
n n
86 Engineering Mathematics - I

∴ By ratio test, Σun converges when x < 1 and diverges when x > 1 and when x = 1
the test fails.
n 1 un
When x = 1, un = ; Taking νn = , Lt = 1 (verify)
n 2 +1 n n →∞ νn
By comparison test, Σun diverges. [Since Σνn diverges by p-series test as p =
1
< 1]
2
Hence Σun converges when x < 1 and diverges when x > 1.
xn
25. Test for convergence the series Σ (x > 0) [JNTU 2007, 2008]
n
Solution
xn x n +1 u ⎛ n ⎞
un = , un+1 = ; Lt n +1 = Lt ⎜ ⎟ x=x
n n + 1 n →∞ un n →∞
⎝ n +1⎠
∴ By comparison test, Σun converges when x < 1 and diverges when x > 1. when x
= 1, the test fails.
1
When x = 1, un = and Σun is divergent (p series test – p = 1)
n
∴ Σun is convergent when x < 1 and divergent when x > 1.
⎛ 1 ⎞
∞
sin ⎜ ⎟
26. Find whether the series Σ (–1) n ⎝ n ⎠ is absolutely convergent or
n=2 (n–1)
conditionally convergent. [JNTU 2006]
Solution
When n > 2, we have,
⎛ 1 ⎞
sin ⎜ ⎟
un = ⎝ n⎠ ; Let νn =
1
;
(n–1) n3/2

un n3/2 ⎡ ⎛ 1 ⎞⎤
∴ = ⎢sin ⎜ ⎟⎥
νn n–1 ⎣ ⎝ n ⎠⎦
Sequences and Series 87

⎛ 1 ⎞
⎜ sin ⎟⎛ n ⎞
⎛u ⎞ n
Lt ⎜ n ⎟ = Lt ⎜ ⎟⎜ ⎟ =1
n →∞
⎝ νn ⎠ n →∞
⎜ 1 ⎟⎝ n −1 ⎠
⎜ ⎟
⎝ n ⎠
∴ By comparison test Σ un and Σνn behave alike. But Σνn is convergent by
p-series test (p = 3/2 > 1).
∴ Σ un is convergent. Hence Σun is absolutely convergent.
∞ cos n π
27. Test whether the series Σ converges absolutely [JNTU 2006]
n =1 n 2 +1

Solution
∞ cos n π ∞ 1
The given series is Σ = Σ (−1) n un where un = 2 .
n =1 2
n +1 1 n +1
It is obvious that u1 > u2 > …. un > un + 1 > …. and Lt un = 0
n →∞

∴ By Leibnitz’s test given series is convergent.

1 1
Σ (−1) n . un = Σ which is convergent (Take ν = 2 and apply comparison
n +1
2
n
test. This is an exercise to the reader)

Hence given series is absolutely convergent.

28. Find whether the following series converges absolutely or conditionally

1 1 1 1.3.5 1.3.5.7
− . + − + ..... [JNTU 2007]
6 6 3 6.8.10 6.8.10.12
1 ⎡ 1 3.5 3.5.7 ⎤
Solution The given series is ⎢1− + − + .....⎥
6 ⎣ 3 8.10 8.10.12 ⎦
1 ⎡ 2 ⎧ 3.5 3.5.7 ⎫⎤
= ⎢ +⎨ − + .....⎬⎥
6 ⎣ 3 ⎩ 8.10 8.10.12 ⎭⎦
3.5 3.5.7
We can take the series as − + ....., (neglecting other terms) = Σun,
8.10 8.10.12
3.5.7.....(2n + 3) 3.5.7.....(2n + 3) (2n + 5)
where un = , so that un +1 =
8.10.12.....(2n + 8) 8.10.12.....(2n + 8) (2n + 10)
88 Engineering Mathematics - I

⎛ u ⎞ ⎡ (2n + 10) ⎤
It can be seen that Lt ⎜ n
n →∞ ⎜ u
⎟⎟ = nLt ⎢ (2n + 5) ⎥ = 1, so that ratio test fails.
⎝ n +1 ⎠
→∞
⎣ ⎦

⎛ u ⎞ ⎡ (2n + 10) ⎤ ⎛ 5n ⎞ 5
Lt ⎜ n − 1 n ⎟ = Lt ⎢ − 1⎥ n = Lt ⎜ ⎟ = >1
n →∞ ⎜ u ⎟ n →∞ (2n + 5)
⎝ n +1 ⎠ ⎣ ⎦ n →∞
⎝ 2n + 5 ⎠ 2

∴ By Raabe’s test, Σ un is convergent. Hence the given series is absolutely

convergent.
( −1) n (x + 2)
29. Test for absolute convergence of the series whose nth term is .
2n + 5
[JNTU 2007]
x+2
Solution: Let given series be Σun. Then un =
2n + 5
⎛ 5 ⎞
2n ⎜1 + n ⎟
x+2 u 2 +5
n
⎝ 2 ⎠
∴ un +1 = n +1
, = so that, n +1 = n +1 =
2 +5 un 2 +5 ⎛ 5 ⎞
2n ⎜ 2 + n ⎟
⎝ 2 ⎠
un +1 1
∴ Lt = <1 ;
n →∞ un 2

∴ By ratio test, Σ un is convergent. Hence the given series is absolutely

convergent.

30. Test whether the following is absolutely convergent or conditionally

convergent.

( )
∞
Σ (−1) n +1 n +1 − n [JNTU 2008]
1

Solution
The given series is Σun where un = (– 1)n+1 ⎡⎣ n + 1 − n ⎤⎦ . It is an alternating

series.
⎡ n +1 − n ⎤ ⎡ n +1 + n ⎤
n +1 − n = ⎣ ⎦ ⎣ ⎦ = 1
= ν n (say),
n +1 + n n +1 + n
Sequences and Series 89

(1) νn > 0 ∀ n ; (2) Lt νn = 0 and (3) νn + 1 > νn ∀ n; since all conditions of

n →∞

Leibnitz’s test are satisfied, the given series is convergent.

1 1
Further, un = ; Taking ν n = , we have,
n +1 + n n
un n 1
Lt = Lt = ≠0
n →∞ νn n →∞ ⎡ 1 ⎤ 2
n ⎢ 1 + + 1⎥
⎣ n ⎦

1
∴ By comparison test, Σ un and Σνn behave alike. But Σνn = Σ is
n
⎛ 1 ⎞
divergent by p-series test ⎜ p = < 1⎟ .
⎝ 2 ⎠

∴ Σ un is divergent. Hence the given series is conditionally convergent

31. Find the interval of convergence of the series,

1 1 1
+ + + ........
1 − x 2(1 − x ) 3(1 − x)3
2

[JNTU 2007, 2008]

Solution: If the nth term of the given series is un,

1 1
un = ; un+1 =
n(1 − x) n
(n + 1) (1 − x) n +1

⎡u ⎤ 1 n(1 − x) n 1 1 1
Lt ⎢ n +1 ⎥ = Lt = Lt . =
n →∞
⎣ un ⎦ n →∞ (n +1) (1 – x)
n +1
1 n →∞ ⎛ 1 ⎞ (1 − x) 1 − x
⎜1 + ⎟
⎝ n⎠

Now, by ratio test:

(i) x = 1 ⇒ the limit is infinite ⇒ Σun is divergent

(ii) x ≠ 0 and x > 1 ⇒ the limit is < 1 ⇒ Σun is convergent

90 Engineering Mathematics - I

(iii) x ≠ 0 and x < 1 and > 0 ⇒ the limit is > 1 ⇒ Σun is divergent
1 1
(iv) If x = 0, the series is 1 + + + ....., which is divergent by p-series test
2 3
(p = 1)
(v) If x < 0, the limit is < 1 ⇒ Σun is convergent by ratio test .
Hence the given series converges for all values of (i) x > 1 and x ≠ 0 and
(ii) x < 0.

∴ The interval of convergence of the given series is (– ∞ , 0) ∪ (1, ∞ ).

Objective Type Questions

∞
1
1. The infinite series ∑n
n =1
2
+1
is

(i) convergent (ii) divergent

(iii) oscillatory (iv) none of these [Ans :(i)]
1+ n
2. The series is
1 + n2
(i) convergent (ii) divergent

(iii) oscillatory (iv) none of these [Ans :(ii)]

1 1 1 1
3. The series − + − + .....
1 1 2 2 3 3 4 4

(i) oscillatory (ii) absolutely convergent

(iii) conditionally convergent (iv) none of these [Ans :(ii)]

1 1 1
4. The series 1 − + − ...... is
2 3 4
(i) oscillatory (ii) divergent

(iii) convergent (iv) none of these [Ans :(iii)]

Sequences and Series 91

x 2 x3 x 4
5. The interval of convergence of the series x − + − + ...... ,is
2 3 4

(i) −∞ < x < ∞ (ii) −1 < x < 2

(iii) −1 < x ≤ 1 (iv) none of these [Ans :(iii)]

1 2 3
6. The series + + + .........∞ is
1.2 3.4 5.6

(i) convergent (ii) divergent

(iii) oscillatory (iv) none of these [Ans :(ii)]

1 2 3
7. The series + + + .........∞ , is
1.3 3.5 5.7

(i) conditionally convergent (ii) convergent

(iii) divergent (iv) none of these [Ans :(iii)]

2 3 4 5
8. The series p
+ p + p + p + ........∞ is convergent if
1 2 3 4

(i) p<2 (ii) p= 2

(iii) p>2 (iv) none of these [Ans :(iii)]

9. The series 6 – 10 + 4 + 6 – 10 + 4 + 6 – 10 + 4 + 6............. ∞ is

(i) convergent (ii) oscillatory

(iii) divergent (iv) none of these [Ans :(ii)]

1 1 1
10. The series + + + ......... is
2.4 4.6 6.8

(i) convergent (ii) divergent

(iii) oscillatory (iv) none of these [Ans :(i)]
92 Engineering Mathematics - I

2. Indicate whether the following statements are true or false:

1
1. The series ∑1+ 2 −n
is convergent . ........................ [False]

n2 + 5
2. The series ∑ 2 is not convergent . ...................... [True]
2n + 7
1 1 1
3. The series + + + ........ is divergent.................... [False]
1 2 3

x3 x5
4. The series x − + − + − ......., converges when −1 ≤ x ≤ 1 .. [True]
3 5
( −1)
n −1

5. The series ∑ n.5n

is absolutely convergent. ........................ [True]

6. The series x + 2 x 2 + 3 x 3 + 4 x 4 + ....∞ is convergent if x > 1........ [False]

x 2 x3
7. The series x + + + ........∞ is divergent if x ≥ 1 ............... [True]
2 3
x 2! 3 3! 3 4! 4
8. The series 1 + + x + 3 x + 4 x + ...... + ∞ is convergent
2 32 4 5

if x > e ........................... [True]

3
−n 2
⎛ 1 ⎞
9. The series ∑ ⎜1 + ⎟ is divergent ............................. [False]
⎝ n⎠
2 3
1 2 ⎛3⎞ ⎛4⎞
10. The series + x + ⎜ ⎟ x 2 + ⎜ ⎟ x3 + ....... is convergent
2 3 ⎝4⎠ ⎝5⎠

if x < 1 . ............................. [True]

11. The series 1 − 2 x + 3 x 2 − 4 x 3 + .....∞ ( x < 1) is divergent............ [False]

Sequences and Series 93

x x2 x3 x4
12. The series − + − + ......∞ is convergent ...... [True]
1 + x 1 + x 2 1 + x3 1 + x 4
sin x sin 2 x sin 3x
13. The series − 3 + 3 + ......∞ converges absolutely...... [True]
13 2 3

( −1)
n −1

14. The series ∑ n

is conditionally convergent. ........................ [True]

3n 2 + 5
15. The series whose nth term is is convergent. [False]
( n + 2)
a

3. Fill in the Blanks:

∞
1. The geometric series ∑ ar
n =1
n −1
converges if___________ . [Ans: | r |< 1 ]

2. If a series of +ve terms ∑u n is convergent, Lt un = ______. .

n →∞
[Ans: 0]

∑{ }
∞
3. 3
n3 + 1 − n is __________. [Ans: convergent.]
n =1

∞
3n3 − 4
4. If ∑ is divergent ,value of p is _________. [Ans: ≤ 4 ]
( n + 5)
p
n =1

2n
⎛ n2 − 2 ⎞
5. The interval of convergence of ∑ un where un = ⎜ 2 ⎟ x , is _______.
⎝n +2⎠

[Ans: −1 < x < 1 ]

6. ∑u n is convergent series of +ve series. Then Lt un1 n is_________.

n →∞
( )
[Ans: < 1 ]
94 Engineering Mathematics - I

7. The series 8 – 12 + 4 + 8 – 12 – 4 + .......is _________. [Ans : Oscillatory ]

⎡ ⎧ un ⎫⎤
8. If un > 0, ∀n and ∑u n is convergent, then Lt ⎢ n ⎨ − 1⎬ ⎥ is _______.
⎣ ⎩ un +1 ⎭ ⎦
n →∞

[Ans: > 1 ]
∞
an
∑ ( −1) a , ( a > 0∀n ) is convergent, then for all values of n ,
n
9. If the series n n
n =1 an +1

is______. [Ans: > 1 ]

− n2
⎛ 1⎞
10. If un = ⎜1 + ⎟ , Lt un1 n = _____________. [Ans: 1 e ]
⎝ n⎠ n →∞