BS Grewal.
Uploaded by
Debmalya PariaBS Grewal.
Uploaded by
Debmalya PariaInfinite Series
1. Introduction. 2. Sequences. 3. S
eries : Convergence. 4. General properties. 5. Series of positive
terms 6. Comparison tests. 7. Integral test. 8. Comparison of ratios. 9. D'Alembert's ratio test. 10. Raabe's
test, Logarithmic test. 11. Cauchy's root test. 12. Alternating series ; Leibnitz's rule. 13. Series of
positive or negative terms. 14. Power series. 15. Convergence of Exponential, Logarithmic and
Binomial series. 16. Procedure for testing a series for convergence. 17. Uniform convergence. 18.
Weierstrass's M-test. 19. Properties of uniformly convergent series. 20. Objective Type of Questions.
9.1 INTRODUCTION
Infinite series occur so frequently in all types of problems that the necessity of studying
their convergence o r divergence is very important. Unless a series employed in an
investigation is convergent, it may lead to absurd conclusions. Hence it is essential that the students
of engineering begin by acquiring an intelligent grasp of this subject.
9.2 SEQUENCES
(1) An ordered set of real numbers, a, Ag, Ag, . .
., a, is called a sequence and is
denoted by (a. ). If the number of terms is unlimited, then the sequence is said to be an infinite
sequence and a, is its general term.
For instance (i) 1, 3, 5, 7, ..., (2n-1), ..., (ii) 1, 1/2, 1/3, ..., 1/n
,
... ,
(iii) 1 ,-1,1,-1, ...,(-1)- 1, ... are infinite sequences. (2) Limit. A sequence is said to
tend to a limit I, if for every e > 0, a value N of n can be found such that la -1
1 < e for n N.
We then write Lt (an) = l or s imply (a.) →l as
n →0.
(3) Convergence. If a s equence (a) h as a finite limit, it is called a convergent
sequence. If ((,) is not co
nvergent, it is said to be divergent.
In the above examples, (ii ) is convergent, while (i) and (ii) are
divergent.
) Bounded sequence. A sequence (a) is said to be bounded, if there exists a
(4
uch that a, <k for every n.
number k s
(5) Monotonic sequence. The sequence (a) is said t o i ncrease steadily or to
decrease steadily according as an +
12 a
, or anul sa,, for all values o . Both increasing and
f n
decreasing sequences are called monotonic sequences.
A monotonic sequence always tends to a limit, finite or infinite. Thus, a
sequence which is monotonic and bounded is c
onvergent.
(6) Convergence, Divergence and Oscillation. If Lt (an) = l is finite
and unique then the sequence is said to be convergent.
365
100
366
HIGHER ENGINEERING MATHEMATICS
If Lt (an) is infinite (+ -), the sequence is said to be divergent.
If Lt (an) is not unique, then ( Q.) is said to be oscillatory.
xamine the f ollowing sequences for convergence :
Example 9.1. E
) an= 21
n2 = 2n ( 2
3n un
(ii) a
n
=
21
(iii) a =
3 + (- 1
) ".
1-2/n S
olution. (i) L
t
-=1/3 which is finite and unique. Hence the sequence (a,) is
3+1/n convergent.
a +
+n
(ii) Lt (2")= do. Hence the sequence (a) is divergent.
(it) Lt (3+(-1)"] = 3 + 1 = 4 when n is even
n
>
= 3-1 = 2, when n is odd i.e
., this sequence doesn't have a unique limit. Hence it
oscillates.
PROBLEMS 9.1
Examine the convergence of the following sequences :
3n-1
2. a =
1 + 2/n
An
3. a, = In + (-191-1
4.2. = sin ?
1+ 2n
5. a
= 1/2
6. a, = 1+(-1)
8. a = 2n.
9.3 SERIES
(1) Def. If u ,, ,, ug .., U
n, . .. be an infinite sequence of real
numbers, then
+ ... + x + ... is called an infinite series. An infinite
Un + u, + Ug
series is denoted by Lu, and the sum of its first n terms is
denoted by se
(2) Convergence, divergence and oscillation of a series.
Consider the infinite series Eun = u, + 4y + ug
+ ... +4, + ... and let
the sum of the first n terms be s = +4, + + ... + un
Clearly, s, is a function of n and as n increases indefinitely three possibilities
arise : (i) Ifs, tends to a finite limit as
, the series Xu, is said to be c onvergent. (i) If s
no , tends to ta as n to, the
series Xu, is said to be divergent. ( iii) I f s, does not tend to a unique
limit as n so, then the series Eu, is said to be oscillatory or non
convergent.
Example 9.2. Examine for convergence the series (i) 1 + 2 +3 + ... +n +
... Oo. (vi) 5- 4-1 +
5- 4-1 + 5-4-1 + ...
Solution. () Here
1) $ = 1+ 2+ 3+ ... +n=
n(n+
A2
+
Lt sp=Lt n(n 1) +0. Hence this series is divergent.
2 (ii) Here
s, = 5-4-1+ 5-4-1+ 5-4-1+...n t erms
= 0,5 or 1 according as the number of terms is 3m, 3m + 1, 3m + 2.
Clearly in this case, s, does not tend to a unique limit. Hence the series is
oscillatory.
1
INFINITE SERIES
367
Examples 9.3. Geometrie series. Show that the series 1 ++ 2 + 73 + ... to (i)
converges if I rl <1, (ii) d
iverges i f r21, a
nd (iii) oscillates i frs-1.
Solution. Let
1 +r+r2 + ... + ph=1 <1, Lt r = 0.
S =
Case I . When
r
2
1
1.- 1-r
so that Lt sn=
1
1- Also
en 1-1 . the series is convergent. Case II. (i) When r>1, Lt p >
.
and
ph-1
1 Also
= so that Lt → os
hen r = 1, then s = 1 + 1 + 1..... +1=n
r-1 -1 1-1 .. the series is divergent. (ii) W
Lts →00 ... The series is divergent. Case III. (i) When r = - 1, then the
series becomes 1-1+1-1+1-1... which is an oscillatory series. (it)
When r<-1, let r=-p so that p> 1. Then r = (-1)" pa
1- 1-(-1)"p" as Lt p" → S = =
1-r Lt Sn → -or + ac according as n is even or odd.
Hence the series oscillates.
and
PROBLEMS 9.2
Examine the following series for convergence :
1 1 1 1 1. 1+= += += +..,
2 4 8
2. 1-1.1
1
1
3. 6 - 10 + 4 + 6 - 10 + 4 + 6 - 10 + 4 + ... ..
+ ... 09.
(V.T.U., 2006)
tres. Each time the ball hits the
1-2 2.3 3.4 5. A ball is dropped from a height h me
ground, it rebounds a distance r times the
distance fallen where 0<r<1. If h = 3 metres and r = 2/3
, find the total distance
travelled by the ball.
9.4 GENERAL PROPERTIES OF SERIES
The truth of the following properties is self-evident and these may be regarded as
axioms :
1. The convergence or divergence of an infinite series remains unaffected by the
addition or removal of a finite number of its terms; f or the sum of these terms
being the finite quantity does not on addition or removal alter the nature of
its sum.
2. If a series in which all the terms are positive is convergent, the series remains
convergent even when some o r all of its terms are negative ; for the sum is
clearly the greatest when all the terms are positive.
h
3. T e convergence or divergence of an infinite series remains unaffected by
multiplying each term by a finite number.
9.5 SERIES OF POSITIVE TERMS
1. An infinite series in which all the terms after some particular terms are
positive, is a positive term series. e.g., -7-5-2 +2 + 7 + 13 + 20 + ... is a
positive term series as all its terms after the third are positive.
2. A series of positive terms either converges or diverges to +
6; for the sum of its
first n terms, omitting the negative terms, tends to either a finite limit or + o.
368
HIGHEA ENGINEEAING
HEMATIOS
MAT
no
3. Necessary condition for convergence. If a positive t erm series L
u, is
convergent, then L
t un= 0.
(P. T
. U
. ,
0
20 9) L et
8. = U+ ug + u
g + ... + u. Since Eu, is given to be
convergent. Lt sn= a finite quantity k ( say).
Also Lt sn-1=k
But
Lt Un = Lt (sn -Sn- 1
)=0.
Un =
sn-Sn-1 Hence the result.
Obs. 1 . It i s i mportant to n
ote that the converse of t his result is
not true.
Consider, for instance, the series 1+
+
+ ... +
+....
Since the term go on
descending,
+...+
1
S = 1 +
12 Lt sp = Lt
Vn
13 .
Thus the series is divergent even though
Lt Uly 3 Lt
tu
Hence L p = 0 i s a necessary b
ut not s ufficient condition f or c onvergence of
non
21 Obs. The above result leads to a simple test
tu
for divergence : If L n 0, the s eries E
u, must b
e
divergent.
96 COMPARISON
TESTS
I. If t wo p
ositive t erm series Eu, a
nd Du, be
such that (1) Ev, converges,
(i i) u
. So, f or a
ll v alues of n, t hen L
u,
also c onverges. P
roof. S
ince Ev, is convergent,
Lt (0; + V+ Vy + ... + Un) = a finite
quantity k ( say) Also since u, sv, U, S U , ---, U,
SUR
Adding, U++ ... + un Sv + Vy + ... +
un
Lt (u, +1, + ... + 0,5 Lt (0, +2, + k.
... + vn) =
nt
Hence the series Su, also converges.
reater than a fixed number m,
Obs. If, however, the relation Su, holds for values of re g
then the first m terms of both the series can be ignored without affecting their convergence or
divergence.
II. If two positive term series Eu, a nd E v, b e
such that: ( i ) Ev, d
i verges,
(ii) u
, 2u, f or all values of n, then E
u also
diverges. I ts proof is similar to that of Test I. I II. L
imit form If two p
ositive t erm series
Eu, and Lu, b e such that
Lt
= f inite q
uantity (+0
), then Łu, a nd E
u,
converge or d iverge t ogether.
Proof. Since Lt 4* = 1, a finite
number (40)
By definition of a limit, there exists a positive number e, however small,
such that
1-1 <E
|
<€
for nam
forn 2
m
INFINITE SERIES
369
Or
e
-En-l<
for
n2m
or
1- E< <lte for n
2m Omitting the first m terms of both the series,
we have
l-ec une <l+€ for
all n Case I. When , is convergent, t hen
... +0.) = k, a
Lt (0, + V, + finite
number
...(2)
22
De
Also from
(1),
€, i.e., u, <(1 + E
«m <l + )v, for
all n.
[By (2)
Lt (uz + ug + ... + u
n)<(1 + €) Lt (0; + V2 + ... +on)
ence Zu, is also convergent. Case II. When EU,
= (1 + €)k H
is divergent, t hen
Lt (0, + vg + ... + vn) 100 Also from
(1) 1- Esm or u.>
(1 - £)u, for all n
Lt (u, + 4y + ... + un)>
(1 - €) Lt (u;
+ v, + ... + Un) Hence Eu, is also divergent.
[By (3)
9.7
INTEGRAL TEST
dx
A positive ter
m series f(1) + f(2) + ... + f(n) + ...,
ecreases as n increases, converges or div erges
where f(n) d
according as the integral
5 f(x)
dx
...(1) is finite or infinite.
The area under the curve y=f(x), between any
two ordinates lies between the set of inscribed and escribed
rectangles formed by ordinates at x = 1, 2, 3, ... as in Fig. 9.1.
Then
n+1 f( 1)+ f( 2) + ... + f( n)2 (**f( x) dx 2|( 2) +
f( 3) + ... + f(n +
1)
$2
1
f(x) dx 28m+
- f(1)
1
2
n
1
n+
X
3
Fig. 9.1
Taking limits as n too, we find from the second
inequality that Lt suis
f( 1).
f(x) dx +
Hence if integral (1) is finite, so is Lts . Similarly, from the first inequality, we see that if the
integral (1) is infinite, so is Lts. But the given series either converges or diverges to + , i.e., Lts, is
either finite or infinite as n +
.
Hence the result follows.
Example 9.4. Test for Comparison. Show that the
p-series
,ne 7P2P
(i) converges for p > 1 (ii) diverges
for p s 1.
(P.T.U., 2009; V
.T.U., 20
06
; Rohtak,
2003)
Solution. By the above test, this series will converge or diverge according as
is finite or infinite.
370
HIGHER ENGINEERING MATHEMATICS
Ifp71,
ALT
P
m
+
1
x
m
. finite
*, i.e
for p > 1
P- 1'
o dac
fo log x + , this proves the result.
r p <
1
Ifp = 1, ox =
Obs. Application of comparison tests. Of all the above tests the limit formi is the most useful.
To apply this comparison test to a given series Eu , the auxiliary series Ev, must be
so chosen that Lt(ul) is non-zero and finite. To do this, we take v. equal to that
term of u, which is of the highest degree in 1/n a
nd the convergence or
divergence of .. is known with the help of the above series.
Example 9.5. Test for convergence the series
2.3
135 1. 2.3.4 3.4.5
(P.T.U., 2009)
(a) 4.7.10+ 7.10.13* 10.13.16*.
(V.T.U., 2010)
(iii) 1 + -
209
2n-1 1 2-1/n Solution. (i) We have un =
in n(n+1)(n + 2) na (1+1/n) (1+2/n) T ake
u = 1/n2, then
Lt –
Lt n =
2-1/n
2-0 n+ Un
n = (1 + 1/n)(1 + 2/n) (1
+0)(1+0)
both Xu, and Eu, converge or diverge
= 2, which is finite and non-zero -
together. But Ev, = X1/n2 is known to be convergent. Hence Zu, is also
convergent.
(ii) Here
**(8 + 143"+ 2/3n 7* (32) 43:3
(3n + 1)(3n + 4)(3n +
7)
3
+
3
+
Taking
-, we find that
n n n) Now since Ev, is divergent, therefore {u, is also divergent.
re
(iii) He
z
- webm
=
mai (an ) sonoring the first term.
1, ignoring the first term.
1
n+
1 v.
(72 + 1)n + = 1/n, we have
Taking
1. Lt =
e
= Lt =
=1.70 1 1+1/n ) + (1+1/n)" Now since Xv, is divergent, therefore Lu, is also
+
divergent.
INFINITE
SERIES
371
est t he c onvergence o
Example 9.6. T f
the s eries :
3
-
1
()
Cu
(V.T. U. ,
2008) ( i)
-
(iii)
(V. T.U
.,
2
000
S)
V2" +1
Solution. (i ) We have
have u = Vn #1) -
Solution. (1) We
VnGo=Vin +1) - Vn
= n [(1 + 1n)1
2-
1].
(Expanding by Binomial
Theorem)
1+
2n
8n?
28n?
2
81
Taking u. = 11 Jn , we
have
Lt
“r =
Lt
which is finite and non-zero.
Un
1>
n 702 8n
IT
1
both Eu, and Eu, converge or diverge
together. But Lun = 21 In is known to be
divergent. Hence Lu, is also divergent.
(ii) When x < 1, comparing the given
series Eu, with Eun = x",
we get
Lt Un = Lt
n +1 But
Xv, is
n Un n " +%* 2 )
hen x >
convergent, so Zu i s also convergent. W
1, comparing zu, with Zw, Ex-", we get
Lt Un Lt (_1__
* = Lt -
=
Lt =
1: 321 → O as n
→]
[:: 72 → O as n
→ 0]
0 Wnn
h=
++-1
n
11--2n
But Ew, is convergent, so Eu, is also
convergent.
1,
When x =
n
Su
U = +=+=+... Do which is divergent.
2 2 2 Hence, Eu, converges for x < 1 and x > 1 but
diverges for x = 1.
1-1/3"
ere
(it) H
V 2h + 1 l
ti) Here
V
1+1/2
--*1-0)",(1:18) Un=
(*)***, we get
Taking
Also since Ev, = pole where r = 13/2 is a geometric series
u, is also divergent.
E
having r> 1, is divergent. :
Example 9.7. Determine the nature of the series :
V2-1 V3- 1
33 -1 -1 51 -
11-1.
1
72
(i)
(log na
(iv)
(p > 0) n(log n)p
(P.T.U.,
2010)
_ ( n +
1) -12 /n[(1
+1/n) – 1/ (n) S
olution. (i) We have u, = "
(n + 2)3 - 1 m3/(1+2/n)3
- 1/n?)
372
HIGHER ENGINEERING
MATHEMATICS
Taking on = mte, we
find that
1/(1+1/n) -
1/ 1nl170
n (1 + 2/n) - 1/ n] Since Ev, is
n+ U
convergent, therefore {u, is also
convergent.
(ii) Here
==
=
) 13 ore
=
3!
3
n
5.no
(i) Here , = sin 4-4 [3 B Taking u, =
s, we have
3! n
51 74
=
51,4
Lt Un = Lt 1--
Since . 3!
...=10 1 Un n
Eu, is convergent, therefore Eu, is also
convergent. (iii) W
e have Lt (log n) -
( log n)?
1/4 , ze.,
or (log n) < n 1/4 no
?
(log n
U
2372
225/4
(:
p= 5
/4 > 1)
onverges by p-series. Hence
Since 1/n5/4 c by
comparison test, Eu, also converges.
(iv) Let
i.e., f(x) is a decreasing
function.
Pim) = ndog mo so that
f(w) Cove 275
ra) = l qoayurt det love 27. ( 7)-- (one per dag
op.com Also 5 rew) dx = Se melhor at lave If p > 1, t hen p
- 1 = k(say) > 0 . f(x) dx = log 2* -10 + (log 21+1 which is
finite
12
Thus by integral test, the given series
converges for p > 1. I fp < 1, then 1-p> 0 and
(log x)l-pwas x 0
f( x) dx => 0. Thus the
given series diverges for p < 1. I f p =
1, then [f() dx = [" dx = log (log x) -
J2 x log 3
1.
Thus the given series diverges for p =
PROBLEMS 9.3
Test the following series for convergence :
1.
1
- ...oo
.T.U. , 2000)
(J.N
2
+
+... 09
3.
1
.
1.2
2. 3.
+2+... 3.4
5.6
(Cochin,
2001)
t
+
=
(P.T.U.,
2009)
*
1.3
3.5
5.7
INFINITE
SERIES
373
T
2PP
2 3.5.7
+ ... 00
.U
(V.T ., 2009
S)
0. 1.3.5
5.7.9
3 4 5 6
itz+27+ 64+
smania, 2000
(O
S)
724+1
12. 5 (n + 1)(n
+ 2)
(J . N
. T
.U
., 2006
S)
13.
Nina +1) – n] (V.T
.U
., 2010;
P.T
.U., 2009)
14.
1963 + 1)
- n]
.T
(P .U
., 2007 ;
Rohtak 2003)
WIWOW!
Ven* + 1) –
Jen* – 11
18. F (+ 1) -
J.N. T.U
.,
2003)
Vn log n
+ 273 -
(n
1
9.8 COMPARISON OF
RATIOS
If Eu, a nd E v, b
e two positive t erm series, t hen Eu, c onverges if ( i) E
u,
converges, a
nd ( ii) from a nd after some particular term,
Let the two series beginning from the particular term be u, +1, +
u, + ... and u,
+1, +1, +... If uzcu, uz Uz...
ų v Ug Up
U2 + 3
113 + ...
then
4, +
H2 + Ug + ... =
u Wuy
Hence, if Ev, converges, Eu, also
converges.
u ya Vg Du '"1* 2
*13*
****
Obs. A more convenient form of the above test to apply is as
follows:
Xu, converges i f ( 1) Eu converges and ( i i) f rom and after a
particular term
om> 11 7+1
, Un + 1
Similarly, Eu, diverges, i f ( ) Ev, d
iverges a
nd (w
)
nd after a particular term
from a
mc m
ili
Un +1
9.9
D'ALEMBERT'S
RATIO TEST* In a
positive t erm series
Lun if
Lt "n+1
= 2
, t hen t he series converges f or a
and d
<1 iverges f or å >
1
.
S CON
1 -
Un
Case I. W
hen Lt "n+1
.
=1<1
*Called after the French mathematician Jean l e-Rond d
' A
lembert (1717-1783), who also made
important contributions to mechanics.
374
HIGHER ENGINEERING
MATHEMATICS
BILLO
Led
1
IIS
un
By definition of a limit, we can find a positive number r
(< 1) such that Pn+1 <r for all n > m
Leaving out the first m terms, let the series
be u, +1, + ug + ... so that "2 <r, 3 <r, 44 <r,... and so on.
Then u, + u, + ug + ....
и це из
442 443 42 44 4g us to
+r+ p2 + pod + ....) u g ug u ug ug u
= , which is finite quantity. Hence Eu, is
convergent.
IS COD
[:
r< 1]
n Lt On+1
Case II. Whe
=2> 1
→
In
By definition of limit, we can find m, such that "+1 2 1
for all n 2 m.
un
Leaving out the first m terms, let the series
be
un + ug + ug +
... so that 2 21, 1321, 14 21 and
so on.
+
Uy + + ug + un
Io Ug
... + Un=u (1+42 + us uz +...
u ta đi
| lu
u, L
+ 1 + 1 + ... to n t erms) = n t (u; + ug +
... + un
> Lt (nu), which tends to infinity. Hence Lu, is
. Ratio test fails when a = 1. Consider, for
divergent. Obs.1
s Eu, = 21/np.
instance, the serie
700
n=
Here
2= Lt = Lt
== Lt =
Un n (n+1) 1 n . (1 +1/n) T
hen for all values of p, 1
1 and diverges for p < 1. Hence A = 1
= 1; whereas 21/nP converges for p >
both for convergence and divergence of Lu, which is absurd.
Oly. 2. It is important to note that this test makes no reference to the magnitude of unlu,
but concerns only with the limit of this ratio.
For instance in the series 1 +*++-+...++ ..., the ratio n+1=1 <1 for all finite values of n, but
tends
2 3 4
72
un
1 to
n+ unity as n o. Hence the Ratio test fails although this series is
divergent.
Practical form of Ratio test. Taking reciprocals, the ratio test
can be stated as follows:
In the positive term series Lu, if Ltr = k, then the series converges for k
> 1 and diverges for ks 1 but fails for k= 1.
Example. 9.8. Test for
convergence the series
(P.T.U., 2005; V.T.U., 2003;
1.S.M., 2001)
n
n
→
U11+1
62
(6) 1***** (>0),
+ ... + =
21 - 2n-17... (x > 0).
2 + 1
009; V.T.U.,
(P.T.U., 2
2004)
Solution. (i) We have
un =
n" (n+1) Vna
and un +1
(n + 2)Non +
1)
INFINITE SERIES
375
1+1)
un_= Lt -
Lt - n
+
+
(n
72
+
2
[1 + 2/n
= Lt
1/n) x 2 = r-2
n
+
1
72
/
1
→
1+1/ n
onverges if x-2 > 1, i.e
Hence Zu c ., for x2 <1 and diverges for x2 >
1, then, u = (n +1) Jm mitz 1+1/0 Taking v,
1. Ifx?
=
- m72, we get Lt = Lt 1+1=1, a finite quantity.
• Both Lu, and Eu, converge or diverge together. But
Lun= 312 is a convergent series.
Eu, is also convergent. Hence the given series
converges if x2 <1 and diverges if x2 > 1.
E
n
Un
n=
1 + 1/n
TIL
-1
1
(ii) Here
Un+
un 1
-2
2
-2 +1
2+1 +1 2h+ 2n +1 -2
L ui n 1
-0 2+0 1 1
.,
by Ratio test, Lu, converges for x > 1 i.e
nUn+1 1+0 2-0 x x Thus
for x < 1 diverges for x > 1. But it fails for x = 1.
When x=1,
= Lt 4
Lt un
2
= Lt -2=170
n6n
+
2
+1
n
+
• Xu, diverges for x = 1. Hence the given series converges for x<
1 and diverges for x 21.
Example 9.9. Discuss the convergence of the series
(V.T.U., 2008 S)
.U., 2010) (1+
(6) įmo (P.T 2+3 +.com
Solution. (i) We have un-inje and un +1* I(n + 1)*
+1P
(n + 1)!
Lt -
un
n!
+ 12 + 1)
(n
Lt OnX
2
227
n
U
+1
72
72
(n + 1)!
ndo
= Lt
+ 1) = e. Lt (n + 1) +
(n
- L1 (1) 2009 - 1:(0:4)*. n = 1)
- 4 (2+4)**-+9)= e. Le (n + 1) + (i) Given series in Iv.
= Here 19 - MA SOFTW=(1+4
7
00
Hence the given series is convergent
(n + 1)
(ii ) Given series is
un
+
Un
n! na
+ 1 ) +
(n
(n + 1 )!
Lt Un = 1 n+ Un + 1
= e, which is > 1. Hence the given series is convergent.
1
N
/
376
HIGHER ENGINEERING MATHEMATICS
Example 9.10. Examine the convergence of t he series :
Te..00
( i i) 1
1
a+ +
1
5+1
1 +2 (Q ) 6+ 1) (26+ 1)
1) ( 2a + 1
+
+ 1
(a ) (2 1) (0+ 1) (26 + 1) (36+1)
a + 1) ( 3a +
+ ...
Solution. (i) Here
.
and u
1+
+1
2
1+
in
1
1 +
n+1)
=
+
3
(1 + +1 = Lt
n7x+*+!
31 +3"
t:
+1 → O as n + ]
= Lt
IR
1+1 +1) Also
+1
= 1 if x > 1. n un
+1 :
1+x/n by Ratio test, Xu, converges for x < 1 and fails for x>1.
=+ = + = + ... +og, which is divergent.
1 1 1 When x = 1, u =
Hence the given series converges for x < 1 and diverges for x 2 1. (ii)
222
Neglecting the first term, we have
1
na +
nu+
1
Lt Um = Lt N6+1 Lt 6+
1/n
_b.
noun +1nna+n Ratio test, Eu, converges for bla >
y
a B
a+1/n 1 or a
<b, and diverges for a >b. When a = b, the series becomes 1 +1 +1
+..., which is divergent. Hence the given series converges for 0 <a <b a
nd
diverges for 0 <b sa.
PROBLEMS 9.4
Test for convergence the following series :
1. XT23
3.
1+
+
+
.T.U., 2006)
(J.N
2
- tas + 10
5
n(n-1) (n - 2
)
5. 1+2+3+4++ ...
2! 3!
(Kurukshetra, 2005)
(Rohtah, 2005)
/
7713
Kerala, 2005)
in +1
9.
"
(P.T.U., 2006)
(Madras, 2000)
2
3.4
2 .4
3.5.6
2.4.6
b+... (V.T.U., 2010) 3.5.7.8
201012. (09+(12) +(3:
12.
3.5.7)
INFINITE SERIES
377
13. 1
12 22 12.22 32 1.3.5 * 1.3.5.7.9
- +...
(Delhi, 2 002)
4 18
4.12 18.27
4.12.20 1 8.27.36
ta i oo
adras, 2
(M 000)
(J . N
. T
. U. , 2006)
3D ***
(2n-1
)
(V. T. U. , 2
004)
3.6.9...
3 5 2.4.7.10... (3n+1) 3n+2 191.1+0(1+a) (1 + 20) (1 + a)(1+20) (1 +
30)
1+B (1+B) (1 + 2B) (1+B) (1 + 2B) (1 +33)
9.10 FURTHER TESTS OF CONVERGENCE
When the Ratio test f ails, w
e apply the following tests :
(1) Raabe's test*. I n
t he positive t erm series Zum, if it n
An
-1
=k,
then t he series c onverges f or k >
1
a
nd diverges for k < 1, but the
1.
test fails for k =
When k > 1, choose a number p s
uch that k>p > 1, and compare Xu, w
ith the
series
which is
No Whi
1.
convergent since p >
. Eu, will converge, i f from and after some term,
Um (n + 1)
or if. Un >1+P. p(
p- 1) ,
In+1 n 2n2
or if,
+... or if, Lt n
QIT
In
Un
1
>
p(p- 1)
2n
=1>
Lt
p-1)
p(
2n
in
1
+1 J
L
i. e., if k > p, which is true. Hence Lu, is convergent.
The other case when k < 1 can be proved similarly.
f Lt
(2) Logarithmic test. In the positive term series Zu, i n
logm
=k,
then the series converges for k>
1, and diverges for k < 1, but the test fails for
k = 1.
Its proof is similar to that of Raabe's test.
Obs. 1. Logarithmic test is a substitute for Raabe's test and should be applied
when either n occurs as an exponent i n u lunt, or evaluation of
Lt becomes
easier on taking logarithm of u, /unt
Obs. 2. If u/u.. does not involve n as an exponent or a logarithm, the series
Lu, diverges.
54.7... (3n +
Example 9.11. Test for convergence the series 6
1) n (va
ton (V.T.U., 2009; P.T.U., 2006 S) 1 . 2.. .n
* 12
2
()
x2n.
In
1
_
-
. 4.7...(3n+
4) n+1
1.2...(n +
1)
1+1 1 3n+4 x
3 + 4/n
x
_4.7... (3n+1)
un
1
+1 1.2...n LtUm-
Un
+1 3x
n
*Called after the Swiss mathematician Joseph Ludwig Raabe ( 1801-1859).
378
HIGHER ENGINEERING
HEMATICS
MAT
Thus by Ratio t est, the series converges for --> 1,
i.e. , for xs
and diverges for
x >
But it fails
for
*** ventes converges for 3: > 1,
1.2., for <<
* and diven
Now
No
w
xpand b
[E y Binomial T
heorem)
--15--+
+...
Lt nl -
which <1.
39n
Thus by Raabe' s iverges. Hence
test, the series d the
given series converges for x < į and
diverges for x 2 j.
( n
um ! [2(n
+ 1)]! *2 (2n + 1) (2n+2) 1 2(2n +
1) 1 (ii)
Here
Un +1 ((n + 1)!) (2n)! **2(n + 1) (n +
1)
2 22n+1 2 12(2+1/n
) 1
-4 n-
2 Thus by Ratio Test, the series converges
tun+in+ 1 + 1/ 7
for x2 < 4 and diverges for x2 >4. But fails for x2 = 4.
n 11
(2
When x2 = 4,
When xo
= 4,
Center - 1) = -( 2
* 2-1) -
--1
=n
Un + 1
)
2n +
2
Un + V (2n+2 Lt
nun-1)= 11
Thus by Raabe's test, t he series diverges. Hence the
given series converges for x2 <4 and diverges for x2 > 4.
Example 9.12. Discuss the convergence of the
series
22,72 333
(P.T.U., 2008; Cochin, 2005 ;
Rohtak, 2003)
l linn
Ca
1)"+ *"
"x" (n +
n"
(n + 1)! (n + " x Lt
19" 2 (1+1/n) Un 1
n + Un +1 ex Thus by Ratio test, the series converg for < lle and diverges for x >
et us try the log-test.
ut it fails for x =1/e. L
1e. B
Now
Un+
1
(1+1/n)"
1
log"n_= log
n log
|1+
- 12
-
Un +1
In
272
2n
3,2+...
POOL
nu
2
Lt n log 4m = 5, which < 1. Thus by the
log-test, the series diverges. Hence the given series
converges for x < 1le a nd diverges for x 2 1le. E xample 9.13. Discuss
the convergence of the hypergeometrie series
a.ß a(a +1) BOB + 1) 2 ala +
1) (a +
2) B(B +
1)(B+2) 1 + x +
. y
-3° + ... (Kurukshetra, 2005) 1
1.2. y(y + 1) 1.2.3
. y(y + 1) (y + 2)
379
INFINITE SERIES
Solution. Neglecting the first term, we have
+ n)(B+ n) Un +1 = Un (n + 1) (y + n)
(a
un 1 (n +
1) (y + n) 1
_
1 (1+1/n
)( 1+y/n) 1 1
.. by Ratio t est, the
= Lt n} e up 1 r a => oa (x + 3) (B + 1) = n + (1+c/ n) ( 1 + 87 n
) * *
series converges for 1/x > 1, i.e ., for x < 1, and diverges for x > 1.
But it fails for x = 1. . . let us try the Raabe's test.
Lt na
S(n + 1) (n + y) ( (n + a) (n + B)
= Lt n n2=
-1% =
Lt n
* I
n(1 + y = 0.- B)+y-aß
n + n(a +B) + aß
(1 + Y-C-B) + (
= Lt
=1+y-a-B
1
1
+ (a +B)
+
Thus the series converges for 1 + y-a-B > 1, i. e., f or y> Q + B and diverges for y<a
t
+ B. But it fails for Y= + B. Since u, lun does not involve n a s an exponent or
a logarithm, the series Xu, diverges for y=a + B.
Hence the series converges for x< 1 and diverges for x > 1. When x = 1, the
series converges for y> a+B and diverges for y Sa+B.
PROBLEMS 9.5
Test the following series for convergence
112**********
(Mumbai, 2009)
1
(V.T.U., 2008
; J.N.T.U., 2003)
23 3.4 4.5
(Raipur, 2005)
1r 1.32, 1.3.5 3
2.4 2.4.6
* + ... . (x >0)
2 2.3 2 2.3.4 3 3**3.5* +3.5.7"
(V.T.U., 2009 S)
100 NX
3.6 2
3.6.9
7.10.13
3.3.6.9.124
7.10.13.16
00
#...
. 10
17
7. 1
1
3
(V.T.U., 2007; Raipur, 2005)
1.3 25 1.3.5 x +2.4.5 +2.4.6.7
+ ... 0. (x > 0)
1.3.5 x . 1.3.5.7.9 26 2.4.6.8*2.4.6.8.10 12"
B. 1* 2.4 8.1+1 x
(Rohtak, 2006 S; Roorkee, 2000)
9. 1 +
+ ... 00 (x > 0)
2!
12.52
12.62.92
19. a+x
(a + 2x)
(a + 3x)
11:
1
2
3!
12. 32 (log 299 + x(log 3) + ** (log 4y + ... 0
(V.T.U., 2000)
ala+1) 2 a( a +
1)(a + 2) 3 0,X > 0). 14.
+ ... (a, b > 1+ 7*+B[b+
1) ** + 3(6
+ 1)(b + 2)
380
HIGHER ENGINEERING
MATHEMATICS
TO
9.11 CAUCHY'S ROOT
TEST*
In a positive series Eup , if Lt (umin
= 2, then the series converges for a <
1, a
nd diverges for à >
1.
Case I . When
Lt (un) Un
= 1 < 1. By definition of a
limit, we can find a positive number ra<r <1) such that
) i n <r for all n >
(u m
,
or u,
<
p for a
ll n > m. S
incer <1, the geometric
series is convergent. Hence, by comparison test, Xu, is also convergent. Case I I. W hen L
t
(u) / n = 1 > 1. By definition of a limit, we can find a number m, such that
(un) I n > 1 for all n > m,
o
r u>
1 for all n > m. Omitting the first m
terms, let the series be u; + ug + ug + .... so that u, > 1, uz > 1, uz > 1 and so
on.
Uy + 4y + uz + ... + Un>n and Lt
y + 4y + ... + un) Hence the series Eu, is
(4
divergent. Obs. Cauchy's r oot test fails when a =
1
.
est for convergence t he
Example 9.14. T
series
(i i) E(log n) - e
n (iii) 2
( 1
+1
/n
)" (P.T
. U. , 2
009;
Kurukshetra, 2 005)
no
Solution. (i) We have u =
n°/3.
n7093
Hence the given series converges by Cauchy's root
test. ( i i) Here
( un)!/n =
(log n) - 2n Lt
Un =
Lt (log n)? = 0 (< 1)
Hence, by Cauchy's root
n →
test, the given series converges. (i i)
Here
Un = (1+1/
/n)-2012
1:
Lt log n =
0]
11/7
tu, une fata,
soy
3/2
sau a fost
in
Lt (u) n = Lt -
(1+1 nyn
which is < 1. Hence the given series is
convergent.
noo
Example 9.15. Discuss the nature of the
following series :
(J.N.T.U.,
2006)
m 3+3++(99**+()*2* =
.6630) (6) 3 (1) 4 19.29
(V.T.U.,
2006)
*See footnote p.
144.
INFINITE
SERIES
381
n
+1
Solution. (i ) A
fter leaving the first term, we
find that
*", so that
n + 2)
Lt
(1+1 / n (u)n
= Lt
n+ 1+27n .
: By Cauchy's root test, the given series converges
for x < 1 and diverges for x > 1.
n+1
When x = 1, 4,
1
n+
When x = 1, .-(0:2)
1+
(1+2+)
1)
n+
nos ei
11
Lt nin=1
n
: Lt u == +0. Since u, does not tend to zero, &u, is
divergent. Thus the given series converges for x < 1 and
diverges for x 2 1. (ii) Here (alt) = 0
t (un)n = Lt 1+1. 1 *= Lt
. L
[*. Lt nin=1]
n in
• The given series converges for x <
1 and diverges for x > 1. When x = 1, u =
4 * * - - [1 + 1)
Taking Taking on L. (4.) = + (1+ 1)* = e 40
and finite. - By comparison test both Eu,
and Lu, behave alike.
1). Eu, also diverges. Hence the
But Ev, = 2 is divergent (: p=
given series converges for x <1 and diverges for x 2 1.
=
= e = 0 and finite.
7000)
n n
+ 1) 2+1
'
n +1
(iii) Here
n+1
(i) Here --- (0 Sa+
(o,)* = (01) "[C+1) -21°+(14(1+4) --"
• Lt (,)o'* = 1- (e –
14--19<1
1:: e> 1]
n 700
Thus the given series
converges.
PROBLEMS 9.6
Discuss the convergence of the following series
:
2.
1
(P.T.U.,
2005)
L
(log n)
3.
(P.T.U.,
2010)
+ ... + (> 0)
2
34
[(2n + 1) x)"
20
72
+1
(V.T.U., 2007)
382
HIGHER ENGINEERING
HEMATICS
MAT
9.12 ALTERNATING
SERIES
(1) Def. A series i n which t he t erms a
re a
lternately positive or negative i s c alled an alternating
series.
g + Ug - U + ... conver g
(2) Leibnitz's series. An alternating s eries un - U es i f
(i ) each term is numerically l ess than its p
receding term, a
nd ( ii) Lt u
n
= 0.
and
or as
Lt u n
70, the given series is oscillatory.
n → The given series is u, -u, +
uz-u, +
... Suppose
Lt un = 0
n → Consider the sum of 2n terms. It
can be written as
82n = (uq- u
n) + (ug- ) + ... +
(uan-1-42n)
San = 4,- (uz - ug) - ( 44 - ug) ... - Uan By virtue of (1),
the expressions within the brackets in (3) and (4) are all positive. . It follows
from (3) that s. is positive and increases with n. Also from (4), we note that
sy always remains less than u Hence s., must tend to a finite limit. Moreover
Lt Sun+1 = Lt (S2n + Ulin +1)= Lt San + 0
Thus Lt sn tends to the same finite limit
whether n is even or odd. Hence the given series is
convergent. When
0, Lt San Lt S2n+1. The given series is
Lt Un
oscillatory.
[by (2)
Example 9.16. Discuss the convergence of
the series
(i) 1 -
2*4*6
5 7 9 W
11
8
...
(ii) 102-103.3 * log 4
logo
(P.T.U.,
2010)
Solution. (i) The terms of the given series are alternately positive and negative;
each term is numerically
less than its preceding term
un = Un
-1 =
2012
s
Also Lt un = Lt (1/ \n) = 0. Hence by Leibnitz's rule, the given
series is convergent. (ii) The terms of the given series are
alternately positive and negative and
3 2n +
2n + 1_ "nun-1" 2n2n - 24nin -
1)
-6 <0 for n > 1.
t. e.,
n
.
2n
=1+0
3 un
2n + <un-1 for n > 1. Also Lt un
= Lt Hence by
Leibnitz's rule, the given series is oscillatory. (iii) T
he
terms of the given series are alternately positive and negative.
., log (n + 2) > log (n + 1)
Also n + 2 >n+1, i.e
i.e.,
log (n + 2)log (n + 1), 1.l.,
Un +1 < Un
and
t
Lt u, = L
1)
" in log (n +
Hence the given series is convergent.
INFINITE
SERIES
383
Example 9.17. E
xamine the character of t he
series
(i )
(- 1
)^-1
2
(-1)" -
tema
(i i)
<x< 1.
2n - 1
n(n- 1)
n =
2
Solution. (i) The terms of the given series are alternately positive and
negative; each term is numerically less than its preceding term.
1 -
Un-1
2n - 1
2n- 3* (2n-1)(n-3)
But
Lt u = L
t -1 = Lt -
which is not zero. n
n - 2n - 1 - 2 - 1/n 2 H
ence the given series is
The terms of the given series are
oscillatory. (ii)
alternately positive and negative
*
*-1 "- l(n - 2)x - n]
un-1-nin-1) (n-1) (n - 2) = n(n-1) (n - 2)
un
"<0 for n22,
Un < Un-1 for n 2 2. Also Lt u, = Lt = 0
no n + n(n -1) Hence
the given series is convergent.
(: 0<x< 1)
i. e
.,
(: 0<x<
1)
PROBLEMS 9.7
Discuss the convergence of the following
series :
1.1- 1-1+.... (P.T.U.,
2009)
(V.T.U.,
2010)
22!
RSO
6
11
16
21 26
2.1--13 +
elhi, 2002
...co. 3. 51" (D )
4. Ž (-19-
5.
-+... (Osmania, 2003) 6. 1-2 3
-45.. 7. 1–2 **** *r**.**, (=<). (Cochin, 2005) & į
se non
. . . . . =0 <<1). 10. (è love
2) - (hopa) (lang ) - (it loss) -
2004
(V.T.U., ; Delhi,
2002)
9.13 SERIES OF POSITIVE AND
NEGATIVE TERMS
The series of positive terms and the alternating series are special types of these
series with arbitrary signs.
Def. (1) If the series of arbitrary terms u; + u, +
such that the series 1 uy | + |
ug + ... + un + ... be
Uz | + | ug 1 + ... + | 4 | +... is convergent, then the
series Xu, is said to be absolutely convergent.
) If | | is divergent but Eu, is convergent, then Eu, is said to b
(2 ec
onditionally
convergent.
For instance, the series
1
+
- ... is absolutely convergent, since the
series
1+*+
... is known to be
convergent,
CONVERSE
384
HIGHER ENGINEERING
MATHEMATICS
Again, since the alternating
series 1 -
+
-
+* - ... is convergent, and the series of
absolute values
+
+
+... is divergent, so the original series is conditionally
convergent.
2
3
4
5
bsolutely c onvergent series is necessarily c onvergent b
Obs. 1. An a ut n
ot
conversely. L et Zu be an absolutely convergent series. Clearly
+ Ele + alig + ... +2, +
...
sl ul+ | 42 | + | Uz ! + ... + 1 url + ... which is known to
be convergent. Hence the series Lu, is also convergent.
Obs. 2. As the series II un | is of positive terms, the tests already established for positive
term series can be applied to examine Su, for its absolute convergence. For instance, Ratio test can be
restated as follows:
The s eries
bsolutely c onvergent if L
is a t
<I,
and i s d
ivergent if
Lt
1! > 1. This t est fails w
hen the limit
is unity.
Example 9.18. Examine t he following s eries for
convergence :
(V.T
. U. , 2006)
6)1+4 +*+-+- +.com (ii) - 1
+ 2) +241 + 2 + 3) - g 1 +
2 + 3 +
4) -
+ ... which is,
evidently
Solution. (i) The series of
absolute terms is convergent.
.: the given series is absolutely convergent and
hence it is convergent.
(ii) Here
=(-13-1 n + 1)2
u, =
( -1 -1
(1+
2+
3 + ... +
n)
=(-1
i
-1 n n +1 )
1 7
-1
n
2(n +
1)
20.1.1)2 =
(-19- 1a, (Say).
Then
ana
i.e.
2 (n + 1)2 (n + 2)2 00:1<4,. Also Lt - Lt n = 0.
Thus by Leibnitz's rule, La, and therefore Iu,
is convergent.
14,1 = 2 minTaking on = ., we
note that
Also
Since v, is divergent, therefore £ u | is
., Łu, is convergent but I | unl
also divergent. i.e
is divergent.
Thus the given series Eu i s conditionally
convergent.
Example 9.19. Test whether the following series are absolutely convergent or
not?
(1)
(-1)n-1
12
2n-1
=
n(log n)
Solution. (i) Given series is
u.
INFINITE SERIES
385
1
= 0
n7 2n-1 = 0
This is an alternating series of which terms go on decreasing
and Lt 4, = Lt
. by Leibnitz's rule, Eu,
converges.
1 1 1 The
series of absolute terms is 1 + + + ...
3 57
Here un = 2n1, Taking
0,= , we have
L# - .14 ( 23"-1)-.,-+0 and finite.
.:by Comparison test, Lu, diverges (: EU,
diverges.
Hence the given series converges and the series of absolute terms diverges,
therefore the given series converges conditionally.
he terms of given series are alternately positive and negative. Also
(ii) T
each term is numerically less than the preceding term and Lt um 1= Lt [1/n
(log n) 1 = 0.
: by Leibnitz's rule, the given series
converges.
Als
o
someopata el logo de 5 – 105
2 = 0 and finite.
i.e
., the series of absolute terms
converges.
Hence, the given series converges
absolutely.
9.14 POWER SERIES
(1) Def. A series of the form a, + 2 x + 2, *2 + ... + a, tem + ... where the a's
are independent of x, is called a power series in x. Such a series may converge for some or all values
of x.
(2) Interval of convergence In
the power series (i), un =
0,*".
Un +1
= Lt "n +1*
—= Lt
un +1
= 1, then by Ratio test, the series (1) converges, when lx is
numerically less than 1, i.e.,
d diverges for other
when | * | <1/1 an
values.
Thus the power series (1) has an interval - 11 < x < 1/1 w
ithin which it converges and
diverges for values of x outside this interval. Such an interval is called the interval of convergence of
the power series.
Example 9.20. State the values of x for which the following
series converge:
02-***
za v**
30-vjet.com Solution. (7) Here u.=1– 12–1 and
+2=(- 18 ***
Un +1
_* and
1
Lt
Lt - n+ 1+1/n
I
= | X
72 +
by Ratio test the given series converges for | x | <1
and diverges for | x | > 1.
en sem
386
HIGHER ENGINEERING MATHEMATICS
Let us examine the series for x = +1.
1 1 1 1 For = 1, the series reduces to 1- -+ + ...
x
2 3 4 5 which is an a
lternating series and is convergent.
1 1 1 1 For x =-1, the series becomes - 1+
1 2 3 4 5 which is a divergent series as can be seen by comparison with p-series
1.
when p =
Hence the given series converges for - 1<x<1.
-
+
-
+
-
+
-+
(i i) Here
zl. =
n(1- x)"
n
-
C
LE
By Ratio test, Eu, converges for
1, i.e., | 1-*|>1
i.e
., f or-1>1-2 >1 or x <
0 and x > 2.
Let us examine the series for x = 0 and x = 2.
1 1 1 For x = 0, the given series becomes 1 +
ES ? * 2*34*** n wa
s
+ +++ ... + which is a d
ivergent harmonic series.
1 1 1 (-1) = 2, the given series becomes - 1 + It
For x is an alternating series
which is convergent by Leibnitz's rule
<
Un u n
-1 for all n a
nd Lt un
= 0.]
..+
-
+
Hence the given series converges for x < 0 and < > 2.
Example 9.21. T est t he s eries
== ... f or a
bsolute c onvergence and c onditional c onvergence.
. U. , 2010)
(V.T
(-1)"
+1
= (-191-1 -
Solution. We have u,
LLLLLLLL
n+ 1 )
(2 n
+ 3)
1)
+1
J(2
n +1)
N00
2+1/n
2 + 3/ n
)
Hence the given s eries is absolutely convergent f or | x | <1 a
nd is
divergent for | *|>1 a
nd the test fails for 1 x 1 = 1.
(- 12-1
1,
Forx=
Since 2n + 1<2n + 3 or (2n + 1)-1/2> (2n + 3)- 1/2
n+
(2 1)
i.e
.,
un > Un
t
. Also Lt
no
= Lt n →
=
= 0
1
(2n +
rgent by Leibnitz's test.
.. the series is conve
.> :). Also Lt. x. = . LtMen + 1 = 0. the +.. has un-
vent) Hon vetlin)
But
INFINITE
SERIES
387
On comparing it with vn = [u, is
divergent. Hence the given series is
conditionally convergent for x = 1. For
x =- 1, the series becomes - ( The
Te * 5+...)
For x=-1, the series becomes
But we have seen that the
series
t... is divergent.
17
Hence, the given series is divergent
when x =- 1.
9.15 (1) CONVERGENCE OF
EXPONENTIAL SERIES
x
The series 1 +
+
+ ... +
+ ... oo is convergent for all
values of x.
. T.U.,
(J.N
2006)
-
- Lt x = 0
Here Lt Un+1 =
Lt |
nunn n n - 1)! non Hence the series
converges, whatever be the value of x. (2)
Convergence of logarithmic
series
The
series x
-
+-18
...cis convergent for
– 1<x51.
IL DET LE
n
nn n+
1 ( 1
n +1
no
1+1/n
(- 1)"+1
+1 Here
n Lt
"n+1 = Lt (1) **
=-* Lt =-* Lt
n
Hence the series converges for ' * [<1 and
diverges for * | > 1. When x = 1, the series being
1 - 2+ - + ..., is convergent. When x =-1, the
series being - (1+*+*+ 4 + ...), is divergent.
Hence the series converges for -1<x1. (3) Convergence of
binomial series
n(n-1)
-1)... (n-r+
n(n 1)
The series 1 + nx +
x + ... + =
converges for | *
|<1.
n(n- 1)... (n-r)
n(n-1)... (n-1+1) =
Her
e
r-land
U,=
(-1)!
n
+1
1+1
ner +1 Lt
= Lt.
X= Lt | rur 00
10 T
--1
1.
* = -x for r> n +
Hence, the series
converges for |*|<1.
PROBLEMS 9.8
nvergenc
e : 6 EV
(-12-1
(-1)=1
E
STER
(Rohtak,
2006 S)
1. Test the following series for conditional convergence :
(1)
Vh sin 2x sin 3x 2. Prove that the series sin x
-... converges
absolutely.
3
3
73 3. Test the following series for conditional
convergence :
( 01-11-1
1_1 1.3 1.3.5 P
2P 3P 4P
2 2.4 2.4.6***
+ ... 00
388
HIGHER ENGINEERING
MATHEMATICS
GE
D? 4. Discuss the absolute convergence of (1)
(Hissar, 2
005
S)
Vas
+1)
Veep
+1)
(38
+1)
5. Find the nature of the series
(V.T.U. ,
2009)
1-2
6. For what values of x are the following series
convergent :
. T
(P .U
., 2009 S;
V.T
.U., 2008)
12
(a)
=
72
7. Find the radius of convergence of the
series
(Calicut,
2005)
8. Prove that
+1
+2
+3+7+
4
5
+... Is a divergent
series.
9. Test the series 1 -
-
212 313 4/4 (1) absolute convergence and (ii)
conditional convergenee.
(V.T. U
L, 2007: Rohtak,
2005)
9.16 PROCEDURE FOR TESTING A SERIES FOR
CONVERGENCE
First see whether the given series is (i) a series with
terms alternately positive and negative;
(i) a series of positive terms
excluding power series ; or (iii) a power series.
For alternating s eries ( i) , apply the Leibnitz's rule (8 9.12).
For s eries ( i i), first find u
, and if possible evaluate Lt u, If Lt u, 70, the series
is divergent. If Lt u, = 0, compare Eu, with 21/ nand apply the comparison tests ($ 9.6).
If the comparison tests are not applicable, apply the Ratio test (8 9.9), If Lt u lu =
1, i.e. , the ratio test fails, apply Raabe's teat (8 9.10). If Raabe's test fails for a similar reason,
apply Logarithmic test ($ 9.10). If this also fails, apply Cauchy's root test (8 9.11).
For the p
ower s eries (ii) , apply the Ratio test as in $ 9 .14. If the Ratio test fails,
examine the series as in c ase (i i) above.
PROBLEMS
9.9
Test the convergence of the following
series :
2 -
2
smania, 1999)
-1 (>0). (O
1-
Nnt1
n=1
4
+1
123 1+12 +213 1 +
1
314
& Livi* 2: V2
3*3*
1+
1
2+
2
3 + 13
nx In+ 1Xn+ 2) (>
0).
02
8.
5272
n
(2n-12
gn
INFINITE
SERIES
389
11.
(3x +
5)" (n
+ 1)!
12.
(x +
2)"
3 n
6-1
9"
log 72
14. (-1)"
sin
nx
12-1 V3
1
14 = 1
1-1 + ..... (V.T.U.,
2003)
16
.
33-1
- in l og n) (log log n)
4-1 53
7
4
9.17 UNIFORM
CONVERGENCE
Let
(x) + ug(x) + .....
= į , (x)
n=1
...
(1)
be an infinite series of functions each of which is
-
., 8, () = (x) + u2(x) + ...
defined in the interval (a,b) . Lets (1) be the sum of its first n terms, i.e
+u,(x)
At some point x = xp, if Lt s. (x) = s(x), then the series (1) is said to
converge to sum s(x,) at that point. This means at x = x, given a positive number E, we can
find a number N such that | s(x,)-S (x ) 1 <£ for n >N
...(2) Evidently N will depend on e but generally it will also depend on x,. Now if we keep the same
s but take some other value x, of x for which (1) is convergent, then we may have to change N
for the inequality (2) to hold. If we wish to approximate the sum s(x) of the series by its partial
sums s (x), we shall require different partial sums at different points of the interval and the
problem will become quite complicated. If, however, we choose an N which is independent of the values
N) approxi mates to s(x) for all values
of x, the problem becomes simpler. Then the partial sum s (c), (nu >
and e is uniform throughout this interval. Thus we have
of x in the interval (a, b)
iformly convergent in the interval (a, b), if for a given € > 0, a
Definition. The series Xu, (x) is said to be un
number N can be found independent of X, such that for every x in the interval (a, b),
s(x)-s, (x) < € for all n > N. E
xample 9.21. Examine the
geometric series 1 +x+m2+ ... +-14...for uniform convergence in the interval ( 1)
Solution. We
have
(x) = 1 + x + x2 +
... + to
1
an
d
s(v) = Lt
isla– 6,6) 1
=
1-2 for 1 x | <1 6*1-" which will be < e, if | x |" <e
(1 - 1)
or
1-1 1- x
Choose N such that l x IN= € (1 - x)
N = log (€ (1 - x))/ l og | * | Evidently N increases with the
increase of | x | and in the interval - Sxs, it assumes a maximum value N = log
(ε/2l og į at x = { for a given €.
Thus | s(x) – $(x) | <ą for all n 2 N' for every value of x in the
interval (1,1). Hence
the geometric series converges
uniformly in the interval (-1,3).
$
1
Obs. The gcometric series though convergent in the interual (1, 1), is not uniformly
convergent in this interval, since we cannot find a fixed number N for everyx in this interval
6. N given by ( ) --> as
1)..
390
HIGHER ENGINEERING
MATHEMATICS
9.18 WEIERSTRASS'S
M-TEST*
A series E u, (x) i s u
niformly convergent in an i nterval (a, b ) , if t here exists
a c onvergent series E M, o f positive c onstants such that u n (x) | SM for all values o f x i n ( a, b) .
Since SM is convergent, therefore, for a given € > 0, we can find a
number N, such that s-S1 <for e very n > N here s = M, + M, + ... + M, +
, w Mn + 1 + ... and
sn= M, + M, + ... + M.
This implies that | Mn+1+Mn +2 + ... <e for
every n > N. Since 1 u(x) | SM,
Un+1(x) | +un+ 2(x) + ... I S un+ z(x)] + | Un+
2(x) + ...
SM + ... <e for every n > N. s(x)-s, (x) | <ɛ for every n > N,
1+M+2+
where s(x) is the sum of the series Eu, (x). Since N d oes not
depend on x, the series Eu, (x) converges uniformly in (a,b). Obs. Eu, (x)
is also absolutely convergent for everyx, since 14,(x) | SM
Example 9.22. Show that the following series converges uniformly in any interval: (1) 5
cos nx (Andhra, 1999) ( ii) - 3.42
values
Solution. (i) cos ne - cos på Is =MQ) for
all values of x. Since M = & converges for
p >1,
Since
converges for p > 1, n= 1 . By M-test, the given series
converges uniformly for all real values of x and p > 1. (ii) For all values of
x, n3 +n4x2 > n3
(=M.). But EM, being p-series with p > 1, is
convergent.
..By M-test, the given series converges uniformly in
any interval. Example 9.23. Examine the following
series for uniform convergence : 6 § sin (nx + x)
2 (P.T.U., 2009)
(ii) E n(n + 2)
(P.T.U. , 2005 S)
2)
non +
na + 2n
Since
Solution. (6) sin (nx + xo) |_| sin
(nx + x) | 1
(1)
10m+2) =
-) for all real x.
(= Mn) for all
real x.
Since 3 M, i s convergent, therefore, by M- test, the given series is
uniformly convergent for
all real values
of x.
(ii) For all real values of x, x2 > 0, i.e.,
n9x220
P + n°a? 2np or met het
my CM) Since Ë M, - is convergent for
p > 1,
i.e.,
IS COD
no
=1
.. by M-test, the given series is uniformly convergent for all real values
of x and p > 1.
* Named after the great German mathematician Karl Weierstrass (1815–1897) who made basic
proximation theory, Differential geometry and Calculus of variations. He
contributions to Calculus, Ap
was also one of the founders of Complex analysis.
INFINITE
SERIES
391
s
Hou
-ZS Lit
e
9.19 PROPERTIES OF UNIFORMLY
CONVERGENT SERIES
I. I f t he s eries Eu, ( x) converges u niformly to s um s ( x) in t he interval (a , b) a
nd
each o f t he functions u , ( x) is continuous in t his i nterval, t hen t he sum s ( x) i s a
lso continuous in (a ,
b).
II. I f the s eries Eu, (x) converges uniformly i n t he interval (a
, b
) a
nd e
ach of t he
functions un( x) i s
hen t he series can be i ntegrated t erm by
term
i. e.,
[ [u(x) + Uz (x) + .. .) dx = 4(x) dx + Luz(x) dx +... I II. If
Eu (x) is a convergent series having continuous derivatives of its terms, and the series Lu ( x)
tiated term by term
converges uniformly, then the series can be differen
4 lu () + 2(x)+ ..] = u(x) +
(x) + ..
pb
Example 9.24. Prove that S
(23 dx = {
n(n + 1)
Solution.
* 1 s 1 for O5xs1
for 0 <x< 1. But SM
is a convergent series.
.. by M-test, the series 2(x"In? ) is uniformly convergent in 0 SxS 1. Also
x"In2 is continuous in this interval.
.. the series (x"In2) can be integrated term by term in the
interval 0 SX S1.
i.e.,
li (23dx = E(L dx) - 2 (71* ** dx) = 2 ->
0+1)
Imp. Obs. There is no relation between absolute and uniform convergence. In fact, a series
may converge absolutely but not uniformly while another series may converge uniformly but not
absolutely.
For instance, the series
1 1 1 *2 +1
= ... can be seen to converge uniformly but not absolutely, while the
series 212 +2 22 +3
3
+
2+1
... can be shown to converge absolutely but not
uniformly.
11.
PROBLEMS 9.10
Test for uniform convergence the
series:
1. Se
22
Nin2
6
sin *
sin 2x
sin 3x
sin 4x
.U., 2003; Andhra,
(P.T
2000)
COS X
cos 30 cos 52
s
in 2x 4. sin se
2
4
sin 3x
313
sin 4x
47
t COS 5
22
2 5 + 10+.+ m2 7. Show that
the series sin në and X
cos no converge uniformly for all real values of 0 if 0
<r<1.
8. Show that
W
9. Prove
that
+ ... converges uniformly in the interval x 20 but not absolutely. 3+2 4+ 2
is uniformly convergent for all real values of x.
is uniformly
ennvere n(1 + nx?)
LS
21 + 7230
392
HIGHER ENGINEERING
MATHEMATICS
10. Examine the following series for uniform
convergence :
)
(a
cos(x 2 +
n2x)
n2 + 2)
=]
11. Show that
()
2
** dx = 1 -3.31*5.51
7.71 ***
271 + ..co;
(w)
sin ne
100=2
2. (2n -
134
mu
9.20 OBJECTIVE TYPE OF
QUESTIONS
PROBLEMS 9.11
Choose t he correct answer or f ill up the blanks i n each o
f t he
following problems:
1. The series
+
+ ... converges if
(a) p>0
(6) p <
1
(c) p
>
1
) psl.
(d
2. The
series
(2x)"
converges
if
)-15x31
(a
(c) -2 <<2
*
IA
2 3 4 5 8. The series 2-2+222
+ ... is
(a) conditionally
convergent
(b) absolutely
convergent (c) divergent
(d) none of
the above. 4. Which one of the following s eries is not
convergent ?
21/2*313 414
+ ..
1 1 1 1 (c)
3 +4
2
+ ...091 (d)
x+2 +3 +4 + ... as where
<1.
5. The sum of the alternating harmonic
series 1 -
+
... is (a) zero
(6)
infinite
(c) log 2 (d) not
defined as the series is not convergent. 6. Let u, be a series of
positive terms. Given that w i s convergent and also
Lt 2+1 exists, then the said limit is
nh
(a) necessarily equal to 1
(1) necessarily greater
than 1 (c) may be equal to 1 or less than 1
(d) necessarily less
than 1.
(c) divergent.
(a) convergent (6) oscillatory
8. 1- tato
8. 1
sta +...is
+ is
(a)
oscillatory
(c) divergent
) conditionally convergent (d) absolutely convergent.
(6
INFINITE SERIES
393
1 1 1 1 1 1 1 ** 22 324252627282 ... is (a) conditionally convergent (c)
oscillatory
) convergent (d) divergent.
(6
10.
un
(α) Στη 1
6) Σ.
,
20
+ 1
n(n + 1)
-1) n=0 11.
(n If Eu is a convergent series of positive terms, then Lt un i s
(a) 1
(6) + 1
(c)0
(d) 0. 12. Geometric series 1 +*+ *2 + ... + -1+ ...co
(a) converges in the interval .......
(6 ) converges uniformly in t he interval ......,
(V.T .U., 2010)
13. The series x-
1
+ ...
converges in the interval ......
2
3
4
14. If Lt
n
=k, then Xu, converges for k ....
15. A sequence (a) is said to be bounded, if there exists a number k such that for
every n ,
a
i s.. 16. The series 2-5 + 3 + 2-5 + 3-5+ ... is....... (Convergent etc.)
2
17. The series 1+
+ ...
converges for ...
i
2!
3!
4!
18. If Lt n n _-1) =k, then u d
iverges for k .......
n (Un+1 19. A sequence which is monotonic and bounded is ........
9
20. The series
+
1.2
3.4
= t...ois...... (Convergent etc.) 5.6
эр 24 Р.
21. The series
+ ...
converges for.
29
22. The series
+ ... - is ....... (Convergent etc.)
23. The series
is... (Convergent etc.)
The series 1 – } (x - ) + 1 6-2. + (-1* 6-27
(x - 2)" + ... Do converges in the interval ......
1
to
26. Is the series
convergent?
26. The exponential series 1 + 2 +
+ ...
is absolutely convergent.
True False)
!
2!
(Convergent divergent/oscillatory)
27. The series
- + ... 00 IS... 1.2 2.3 3.4 n( n+1) 28. Is the series In t an 1/n
convergent ?
29. The series
converges for x ......
30. The series
converges uniformly when x lies in the interval .....
394
HIGHER ENGINEERING
MATHEMATICS
(V.7
.U.
2009)
31. Every absolutely convergent series is
necessarily
(a) divergent
)
(6
convergent (c) conditionally convergent
(a) none of
these.
32. The convergence of the
series 1-
2
... is tested
by
3 5
7 (a) Ratio test
(6) Raabe's test (c) Leibnitz's
(d) Cauchy root
test.
.U. ,
(V.T
2009)
33. The series
>
+ 12
VV > O is
(a)
divergent
(b)
convergent
(c)
oscillatory
d) none of
these.
(V.T. U. ,
2010)
(c) oscillatory
(d) none of
these.
(a) convergent (b) divergent
Š 1_ is
convergent.
(True or
False)
2 (log n)
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