Lecture 8

1 Infinite Series
1.1 Definitions & convergence
Definition 1.1.1. 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 series.
k=1

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.

Examples 1.1.2.
∞ 1
xn converges to
P
1) If 0 < x < 1, then .
n=0 1−x
n
xk . Here
P
Solution. Let us consider the sequence of partial sums {sn }, where sn =
k=0

n
X 1 − xn+1 1 xn+1
sn = xk = = − , n ∈ N.
1−x 1−x 1−x
k=0

1 1
As, 0 < x < 1, xn+1 → 0 as n → ∞. Hence sn →
P n
. Thus x converges to . ///
1−x 1−x

P∞ 1
2) The series diverges.
n=1 n
n
P 1
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
 Suppose s2n > (n + 2)/2, then
2 n
X 1
s2n+1 = s2n +
2n + k
k=1
2n
n+2 X 1
> +
2 2n+1
k=1
n+2 2n (n + 1) + 2
= + n+1 = .
2 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 1
3) (Telescopic series:) Show that the series converges to 1.
n=1 n(n + 1)

Solution. Consider the sequence of partial sums {sn }. Then

n k  
X 1 X 1 1 1
sn = = − =1− → 1.
k(k + 1) k k+1 n+1
k=1 k=1

Summarizing this observation, one has the following theorem on Telescopic series

Theorem 1.1.3. Suppose {an } is a sequence of real numbers such that an → L. Then the series
P
(an − an+1 ) converges to a1 − L.

Lemma 1.1.4.
∞
P ∞
P ∞
P
1) If an converges to L and bn converges to M, then the series (an + bn ) converges to
n=1 n=1 n=1
L + M.
∞
P ∞
P
2) If an converges to L and if c ∈ R, then the series can converges to cL.
n=1 n=1
∞
P
Lemma 1.1.5. 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. ///
∞
xn diverges.
P
Example 1.1.6. If x > 1, then the series
n=1
∞
xn converges. Then the nth term, i.e., xn → 0.
P
Solution. Assume to the contrary that the series
n=1
But as x > 1, xn ≥ 1 for all n ∈ N and hence lim xn ≥ 1, which is a contradiction. Hence the series
n→∞
∞
xn diverges.
P
///
n=1

2
 As a first result we have the following comparison theorem:
P
Theorem 1.1.7. Let {an }, {bn } be sequences of positive reals such that an ≤ bn . If bn converges
P P P
then an converges. Also, if an diverges then bn diverges.
P P
Proof. Let sn = a1 + a2 + .... + an and tn = b1 + b2 + .... + bn be the partial sum of an , bn
P
respectively. Then sn ≤ tn . Since bn converges, we have {tn } converges and is bounded. Now since
{sn } is monotonically increasing sequence that is bounded above, we get the convergence of {sn } and
P
hence the convergence of an .
P
If an diverges then sn → ∞. Then tn ≥ sn implies tn diverges to infinity. ///
Examples 1.1.8.
X 1 X n
(a) (b)
2n + n n2 − sin2 n
1
For (a), note that 2n + n > 2n and converges. For (b) note that n2 − sin2 n < n2 and the series
P
P1 2b
n diverges.