Limit of an Infinite Series

Institute of Mathematics
University of the Philippines-Diliman

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Infinite Series

Definition
1. An infinite series is a formal sum
∞
X
xn = x1 + x2 + x3 · · · + xk + · · · , (1)
n=1

of an infinite sequence of numbers xn , called the terms of the

series.
n
X
2. For n = 1, 2, 3, . . ., the expression sn = xk is called the n-th
k=1
partial sum of the series.
3. The series (1) is said to converge if lim sn < ∞. Otherwise, the
series is divergent.

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Infinite Series

4. The number s is called the sum of the infinite series and we write
∞
X
xn = s.
n=1

Note that (i) a series is a result of an infinite addition and (ii) each
partial sum is the sum of finitely many terms only, thus, partial sums
form a sequence.
Other definition:
An infinite series (or simply series) generated by X = {xn } is the
sequence S = {sn } defined by

s1 = x1 , s2 = s1 + x2 , ... sn = sn−1 + xn ,

and so on.

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Infinite Series

Theorem
P P
Let the series xn and yn both converge. Then,
P P P P
1. (xn ± yn ) converges and (xn ± yn ) = xn ± yn , and
P P P
2. (cxn ) converge with (cxn ) = c xn , c ∈ R.

Assignment: Prove item 1 of the previous theorem by CCSeries.

Lemma (Term Test)

P
If xn converges in R, then lim xn = 0.

Proof: P
Suppose xn converges. Thus, lim sn exists. Since xk = sk − sk−1 ,
then lim xk = lim(sk − sk−1 ) = lim sk − lim sk−1 = 0. 2

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Infinite Series
Remarks:
1 The previous lemma
P provides a simple test for divergence, i.e., if
lim xn 6= 0 then xn diverges.
2 If lim xn = 0, the series may or may not converge.
Examples:
∞
X n n 1
1. The series is divergent since lim = 6= 0 as
2n + 1 2n + 1 2
n=1
n → ∞ so it fails the Term Test.
∞
X n n
2. Consider the series 2
. Note that lim 2 = 0 as
n +1 n +1
n=1
n → ∞ and the series diverges.
∞
X 1 1
3. Consider the series . One has lim 2 = 0 as
n2
+n n +n
n=1
n → ∞. Unlike in example 2, this series converges.
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Infinite Series
Theorem
P
Let {xn } be a sequence of nonnegative real numbers. Then xn
converges iff the sequence S = {sk } of partial sums is bounded. In this
case, X
xn = lim sk = sup{sk }.

Proof:
Since 0 ≤ xn , the sequence of partial sums is monotone increasing.
By the Monotone Convergence Theorem, the sequence S converges iff
it is bounded. 2
Cauchy Criterion for Series (CCSeries)
P
The series xn in R converges iff ∀ε > 0, ∃M(ε) ∈ N s.t. if
m > n ≥ M(ε),

|sm − sn | = |xn+1 + xn+2 + · · · + xm | < ε.

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Infinite Series

P
Intuitively, this criterion says that, xn converges iff arbitrarily long
sums
xn+1 + xn+2 + · · · + xm , m > n,
can be made as small as we please by choosing n large enough.

Definition
P
Let {xn } be
Pa sequence in R. The series xn converges absolutely if
the series |xn | is convergent in R. A series converges conditionally,
if it converges but not absolutely.

Note: if the elements of the series are nonnegative real numbers, there
is no distinction between ordinary convergence and absolute
convergence.

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Infinite Series

Theorem
P
If a series xn in R is absolutely convergent, then it is convergent.

Proof:P
Since |xn | converges, it follows from CCSeries that given ε > 0,
there exists M > 0 s.t. whenever m > n ≥ M,

||xn+1 | + |xn+2 | + · · · + |xm || = |xn+1 | + |xn+2 | + · · · + |xm | < ε.

From the triangle inequality, one has

|xn+1 + xn+2 + · · · + xm | ≤ |xn+1 | + |xn+2 | + · · · + |xm | < ε,

2
P
so that, again, by the CCSeries, xn is convergent.

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Examples

Examples:
∞
X
1. Consider the geometric series given by an . A necessary
n=0
condition for its convergence is that lim an = 0. This means, we
should have |a| < 1. If m > n ≥ M for some M ∈ N,

an+1 − am+1
sm − sn = an+1 + an+2 + · · · + am = . (2)
1−a
Hence, triangle inequality yields

|an+1 | + |am+1 |
|an+1 + an+2 + · · · + am | ≤ , m > n.
|1 − a|

If |a| < 1, |an+1 | → 0. Thus, CCFSeries implies that the

geometric series is convergent for |a| < 1.

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Examples

If otherwise, then lim an 6= 0 and so the series diverges by the

simple test for divergence.
To find the sum, we let n = 0 in (2). Taking the limit as m → ∞,
1
next, gives the value . Thus,
1−a
∞
X 1
an = for |a| < 1.
1−a
n=0

∞
X 1
2. Take the harmonic which is known to diverge. The n-th
n
n=1
partial sum is
1 1 1
sn = 1 + + + ··· + .
2 3 n

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Examples

Consider the following subsequence from {sn }:

s1 = 1
1
s2 = 1 + 2
1 1 1


s4 = 1 + 2 + 3 + 4
1 1 1 2


≥ 1+ 2 + 4 + 4 =1+ 2
1 1 1 1 1 1 1



s8 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
1 1 1 1 1 1 1



≥ 1+ 2 + 4 + 4 + 8 + 8 + 8 + 8
3
= 1+ 2

and so on. In general, s2k ≥ 1 + 2k for all k. This means, the

subsequence {s2k } is unbounded which makes the sequence {sn }
diverge.

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Examples
X 1
3. Consider the p-series where 0 < p ≤ 1. Note that for
p
np
these values of p, n ≤ n for n ∈ N. Thus,
1 1
≤ p.
n n
Since the partial sums of the harmonic series are unbounded, this
inequality shows that the partial sums of the p-series where
0 < p ≤ 1 is also not bounded. Therefore, this series diverges for
these values of p.
4. Consider the p-series for p > 1. Note that the partial sums are
monotone so that to prove that is it convergent, it is sufficient to
show that some subsequence remains bounded.
Let kr = 2r − 1 and consider the subsequence skr , r ∈ N. One
has k1 = 21 − 1 gives sk1 = s1 = 1. If k2 = 22 − 1 = 3,

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Examples

 
1 1 1 2 1
sk2 = s3 = + p
+ p <1+ p
= 1 + p−1 ,
1 2 3 2 2
and when k3 = 23 − 1 = 7,
 
1 1 1 1 4
sk3 = s7 = sk2 + p
+ p + p + p < sk2 + p
4 5 6 7 4

1 1
<1+ + .
2p−1 4p−1
1
Let a = 2p−1 . Since p > 1, we have 0 < a < 1. It can be shown
by mathematical induction that if kr = 2r − 1, then

0 < skr ≤ 1 + a + a2 + a3 + · · · + ar−1 .

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Examples

1
This means that the number 1−a is an upper bound for the partial
sums of this series for p > 1, that is, the sequence of the partial
sums is bounded. Therefore, by a previous theorem, the series is
convergent for p > 1.
X 1
5. Consider the series . Using partial fractions,
n2 + n
1 1 1 1
= = − .
n2 +n n(n + 1) n n+1

Thus,
     
1 1 1 1 1 1
sn = 1 − + − +···+ − = 1− .
2 2 3 n n+1 n+1

Therefore, the series is convergent to 1.

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