Series
(Lecture 10 − 12)
Engineering Calculus
School of Engineering and Applied Sciences
Department of Mathematics
Bennett University
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 1 / 18
Convergence
Definition
Let {an } be a sequence of real numbers.
(a) An expression of the form
a1 + a2 + . . . + an + . . .
is called an infinite series.
(b) The number an is called as the nth term of the series.
n
P
(c) The sequence {sn }, defined by sn = ak , is called the sequence of partial sums of the
k=1
series.
(d) If the sequence of partial sums converges to a limit L, we say that the series converges and
its sum is L.
(e) If the sequence of partial sums does not converge, we say that the series diverges.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 1 / 18
Convergence
Example
∞ 1
xn converges to
P
If 0 < x < 1, then .
n=0 1−x
n−1
xk . Here
P
Solution: Let us consider the sequence of partial sums {sn }, where sn =
k=0
n−1
X 1 − xn 1 xn
sn = xk = = − , n ∈ N.
k=0
1−x 1−x 1−x
1
As, 0 < x < 1, xn → 0 as n → ∞. Hence sn → . Thus the given series converges to
1−x
1
.
1−x
Example
∞
log( n+1
P
The series n
) diverges.
n=1
Hint: Sn = log(n + 1) → ∞ as n → ∞.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 2 / 18
Convergence
Example (Telescopic series)
∞
P 1
Show that the series converges to 1.
n=1 n(n + 1)
Solution: Consider the sequence of partial sums {sn }. Then
n n
X 1 X 1 1 1
sn = = − =1− → 1.
k=1
k(k + 1) k=1
k k + 1 n + 1
∞
P 1
Thus the series converges and it converges to 1.
n=1 n(n + 1)
Theorem on Telescopic series
Suppose {an } is a sequence of non-negative real numbers such that an → L. Then the series
P
(an − an+1 ) converges to a1 − L.
Proof: Let Pn be the sequence of partial sum of the series bn = an − an+1 . Then
Pn = a1 − an → a1 − L as n → ∞.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 3 / 18
Convergence
Lemma
∞
P ∞
P ∞
P
(a) If an converges to L and bn converges to M, then (an + bn ) converges to L + M.
n=1 n=1 n=1
P∞ ∞
P
(b) If an converges to L and if c ∈ R, then the series can converges to cL.
n=1 n=1
Theorem (Necessary condition for convergence)
∞
P
If an converges, then lim an = 0.
n=1 n→∞
∞
P
Proof: Suppose an = L. Then the sequence of partial sums {sn } also converges to L. Now
n=1
an = sn − sn−1 → L − L = 0 as n → ∞.
∞
P
• It is possible that an → 0 and an diverges.
n=1
∞
log( n+1 ) diverges, however log( n+1
P
Example: n n
) → 0.
n=1
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 4 / 18
Convergence
∞
P
• If lim an 6= 0. Then an diverges.
n→∞ n=1
∞
xn diverges.
P
Example: If x > 1, then the series
n=0
∞
n
x converges. Then xn → 0. But as x > 1, xn ≥ 1 for all
P
Solution: Assume that the series
n=0
∞
n ∈ N and hence lim xn ≥ 1, which is a contradiction. Hence the series xn diverges.
P
n→∞ n=1
Theorem (Necessary and sufficient condition for convergence)
∞
P
Suppose an ≥ 0 for all n. Then an converges if and only if {sn } is bounded above.
n=1
Proof: Note that under the hypothesis {sn } is an increasing sequence.
Example
P∞ 1
The series diverges.
n=1 n
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 5 / 18
n
1
P
Solution: Consider the sequence of partial sums {sn }, where sn = k
. Now, let us examine
k=1
the subsequence s2n of {sn }. Here
s2 = 1 + 1/2 = 3/2,
s4 = 1 + 1/2 + 1/3 + 1/4 > 3/2 + 1/4 + 1/4 = 2,
1 1 1 1 n
s2n ≥ 1 + + 2. + 4. + ... + 2n−1 . n = 1 + .
2 4 8 2 2
Thus the subsequence {s2n } is not bounded above and as it is also increasing, it diverges. Hence
P∞ 1
the sequence diverges, i.e., the series diverges.
n=1 n
∞
P P∞
Remark: Note that an converges if and only if an converges for any p ≥ 1.
n=1 n=p
Theorem (Comparison Test)
Let {an }, {bn } be sequences of positive reals such that an ≤ bn for n ≥ k for some k. Then
P P
1 If bn converges then an converges.
P P
2 If an diverges then bn diverges.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 6 / 18
Examples
∞ ∞
1 1 11
P P
1 The series (n+1)2
converges, because (n+1)2
≤ n2n2
and
converges.
n=1 n=1
∞ ∞
1
converges, because 2n2 − n ≥ n2 and 1
P P
2 The series 2n2 −n n2
converges.
n=1 n=1
∞ ∞
1√
diverges, because n+1√n ≥ 2n1 and 1
P P
3 The series n+ n n
diverges.
n=1 n=1
P∞ 7 7 1 1 P∞ 1
4 The series =
diverges, because ≥ and diverges.
n=1 7n −2 7n − 2 n − 2/7 n n=1 n
P∞ 1 1 1 P∞ 1
5 The series converges, because ≤ n and n
converges.
n=0 n! n! 2 n=0 2
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 7 / 18
Theorem (Cauchy condensation test)
∞
P
Let {an } be an decreasing sequence of positive numbers. Then an converges if and only if
n=1
∞
2n a2n converges.
P
n=0
Examples
∞ 1 ∞ 1 ∞ 1
2n n p =
P P P
(1) Consider the series p
, p > 0. Then we have n )p−1
which
n=1 n n=1 (2 ) n=1 (2
converges for p > 1 and diverges for p ≤ 1.
∞ 1 ∞ 1 1 P ∞ 1
2n n
P P
(2) Consider the series . Here n
= which diverges. Hence
n=2 n log n n=2 2 log 2 log 2 n=2 n
the given series diverges.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 8 / 18
Theorem (Limit comparison test)
Let {an } and {bn } be two sequences of positive numbers. Then
an P∞ P∞
(a) if lim = c > 0, an and bn both converge or diverge together.
n→∞ bn n=1 n=1
an P∞ P∞
(b) if lim = 0 and bn converges, then an converges.
n→∞ bn n=1 n=1
an ∞
P ∞
P
(c) if lim = ∞ and bn diverges, then an diverges.
n→∞ bn n=1 n=1
Example
P∞ 2n + 1 2n + 1 1
(1) Consider the series 2
. Here an = 2
. Let bn = . Then
n=1 (n + 1) (n + 1) n
2n + 1
an (n + 1)2 2n2 + n P∞
1
= = 2 → 2 as n → ∞. Further, n
diverges. Thus by
bn 1 n + 2n + 1 n=1
n
limit comparison test, the given series diverges.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 9 / 18
Examples
1∞
P 1 1
(2) The series n −1
converges. Here an = n . Let bn = n . Then
n=1 2 2 − 1 2
an 2n P∞ 1
= n → 1. Further, converges and hence the given series converges.
bn 2 −1 n=1 2
n
∞ −n −n
e
converges. Here an = en2 and bn = n12 . Then abnn = e−n → 0 as n → ∞.
P
(3) The series n2
Pn=11
Further, n2
converges and hence the given series converges.
∞
1
log(1 + n1 ) converges. Take bn = n12 . Then lim abnn = 1.
P
(4) The series n
n=1 n→∞
∞
1 1 an
P
(5) The series sin n2
converges. Take bn = n2
. Then bn
= 1.
n=1
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 10 / 18
Definition (Absolute convergence)
∞
P ∞
P ∞
P
(a) Let an be a series of real numbers. If |an | converges, we say that an converges
n=1 n=1 n=1
absolutely.
∞
P ∞
P ∞
P
(b) If an converges but |an | diverges, we say that an converges conditionally.
n=1 n=1 n=1
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 11 / 18
Examples
P∞ (−1)n
(1) The series converges absolutely.
n=1 n!
∞
P (−1)n
(2) The series n2
converges absolutely.
n=1
P∞ (−1)n+1
(3) The series converges conditionally.
n=1 n
P∞ (−1)2n−1
(4) The series converges conditionally.
n=1 2n − 1
Theorem
∞
P ∞
P
If an converges absolutely, then an converges.
n=1 n=1
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 12 / 18
Tests for absolute convergence
Theorem (Comparison test)
∞
P ∞
P
Let an be a series of real numbers. Then, an converges absolutely if there is an absolutely
n=1 n=1
∞
P
convergent series cn with |an | ≤ |cn | for all n ≥ N, N ∈ N.
n=1
Theorem (Ratio test)
∞
P
Let an be a series of real numbers. Let
n=1
an+1
L = lim
.
n→∞ an
Then
∞
P
(a) an converges absolutely if L < 1.
n=1
P∞
(b) an diverges if L > 1.
n=1
(c) the test fails if L = 1.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 13 / 18
Examples
P∞ nn
(a) The series diverges.
n=1 n!
Here n n
(n + 1)n+1 n!
an+1 n+1 1
= = = 1+ → e,
an (n + 1)! nn n n
which is greater than 1. So L > 1. Thus the given series diverges.
P∞ xn
(b) For every x ∈ R, the series converges.
n=0 n!
Here
an+1 xn+1 n! x
= = → 0.
an (n + 1)! xn n+1
Therefore L = 0 < 1. Thus, for all x ∈ R, the given series converges.
Theorem (Root test)
∞
P p
n
Let an be a series of real numbers. Let L = lim |an |. Then
n=1 n→∞
(a) the series converges absolutely if L < 1;
(b) the series diverges if L > 1;
(c) the test fails if L = 1.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 14 / 18
Examples
P∞ xn
(1) Find the value of x ∈ R for which the series converges or diverges.
s n=1 n
xn xn x
Here an = . Therefore, n = √ n
→ |x|. Thus the series converges for |x| < 1
n n n
and diverges for |x| > 1.
P∞ xn
(2) Find the value of x ∈ R for which the series n
converges.
n=1 n
n
x x
p
Here an = n . Then, n |an | = → 0. Thus the series converges for x ∈ R.
n n
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 15 / 18
Alternating series
Definition
∞
(−1)n+1 an is an
P
An alternating series is an infinite series whose terms alternate in sign. i.e.
n=1
alternating series.
Theorem (Leibniz’s test)
Suppose {an } is a sequence of positive numbers such that
(a) an ≥ an+1 for all n ∈ N and
(b) lim an = 0,
n→∞
∞
(−1)n+1 an converges.
P
then the alternating series
n=1
Example
∞
(1 − 21/n )(−1)n+1 .
P
Consider the series
n=1
Here an = 1 − 21/n → 0 as n → ∞. Also an ≥ an+1 for all n ∈ N. Hence the series converges.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 16 / 18
Examples
∞ (−1)n+1
P
(a) Consider the series .
n=1 n
Then an ’s of this series satisfies the hypothesis of the above theorem and hence the series
converges.
∞
(−1)n+1 log1 n .
P
(b) Consider the series
n=2
1
Then an = log n
satisfy the hypothesis of the above theorem and hence the series converges.
Result
(a) Grouping of terms of a convergent series does not change the convergence and the sum.
However, a divergent series can become convergent after grouping of terms.
(b) Rearrangement of terms does not change the convergence and the sum of an absolutely
convergent series.
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 17 / 18
Lecture 10 − 12 (EMAT101L - Engineering Calculus) Series 18 / 18
