Chapter 1.

Infinite Series and Convergence

Definition 1.1 (Infinite Series). An infinite series is an expression of the form

∞
X
ai = a1 + a2 + a3 + · · · + an + . . . ,
i=1

where a1 , a2 , a3 , . . . are called the terms of the series.

Remark 1.2. A series is the sum of the terms of a sequence.

1.0.1 Sequence of Partial Sums and Convergence of a Series

P∞
For the series i=1 ai = a1 + a2 + a3 + · · · + an + . . . , the partial sums are defined as:
n
X
s 1 = a1 , s2 = a1 + a2 , s3 = a1 + a2 + a3 , ..., sn = ai .
i=1

The sequence of partial sums is {s1 , s2 , s3 , . . . , sn , . . . }.

Definition
P∞ 1.3 (Convergence of a Series). Let {sn }∞ n=1 be the sequence of partial sums of
i=1 ai . If (sn ) converges to a limit S, then the series converges and S is called the sum of
the series, i.e.,
X∞
S= ai .
i=1

Note. If the limit of the sequence of partial sums (sn ) does not exist, then the series diverges.

Example 1.4. Find the sum of the series

1 1 1
+ + + ...
2 4 8
if it exists.

1
Chapter 1: Infinite Series and Convergence

Solution. The partial sums are

1 1 1 3 1 1 1 1
s1 = , s2 = + = , ..., sn = + + ··· + n = 1 − n.
2 2 4 4 2 4 2 2
Taking the limit as n → ∞, we have

S = lim sn = 1,
n→∞

so the series converges and

1 1 1
+ + + · · · = 1.
2 4 8
Example 1.5. Determine if the series
∞
X
(−1)n+1 = 1 − 1 + 1 − 1 + 1 − 1 + . . .
n=1

converges or diverges. If it converges, find its value.

Solution. The partial sums are

s1 = 1, s2 = 0, s3 = 1, s4 = 0, . . .

and the sequence of partial sums is

(sn ) = {1, 0, 1, 0, 1, 0, . . . }.

Since the sequence of partial sums does not converge (the two subsequences s2k and s2k−1
do not converge to the same limit), the series
∞
X
(−1)n+1
n=1

diverges.
Example 1.6. Determine if the series
∞
X 1
k=1
k(k + 1)

converges or diverges. If it converges, find its value.

Solution. The partial sums are
1 1 1 2
s1 = , s2 = + = .
2 2 6 3
The n-th partial sum is
n
X 1 1
sn = =1− ,
k=1
k(k + 1) n+1

2 KPIS. Kimuli
where we used
1 1 1
= −
k(k + 1) k k+1
to write it as a telescoping sum. Therefore,

S = lim sn = 1,
n→∞

implying that
∞
X 1
k=1
k(k + 1)
converges and its value is 1.

Example 1.7. Determine if the series

∞
X 1
k=1
(k + 2)(k + 3)

converges or diverges. If it converges, find its value.

Solution. Using the same telescoping method as in Example 3.3, the series converges and
its value is
1
.
3
Theorem 1.8. Let ∞
P P∞
n=1 an and n=1 bn be infinite series of real numbers, and let c ∈ R.
Then:
∞
X ∞
X
(a) can = c an
n=1 n=1

∞
X ∞
X ∞
X
(b) an ± bn = (an ± bn )
n=1 n=1 n=1
P∞
Proof. (a) Consider the sequence of partial sums (sn ) of n=1 an , where
n
X
sn = ak .
k=1
P∞
By hypothesis, sn → s as n → ∞. The n-th partial sum for the series n=1 can is
n
X
cak = ca1 + ca2 + · · · + can = c(sn ).
k=1

Since
lim (csn ) = c lim sn = cs,
n→∞ n→∞

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Chapter 1: Infinite Series and Convergence

the sequence of partial sums (csn ) converges to cs, implying

∞
X ∞
X
can = c an .
n=1 n=1
P∞ P∞
(b) Consider the sequences of partial sums (sn ) and (tn ) of n=1 an and n=1 bn , re-
spectively:
Xn Xn
sn = ak , tn = bk .
k=1 k=1
P∞
The n-th partial sum of n=1 (an + bn ) is
n
X
(ak + bk ) = (a1 + b1 ) + (a2 + b2 ) + · · · + (an + bn ) = sn + tn .
k=1

Since
lim (sn + tn ) = lim sn + lim tn = s + t,
n→∞ n→∞ n→∞

the sequence of partial sums (sn + tn ) converges to s + t, implying

∞
X ∞
X ∞
X
(an + bn ) = an + bn .
n=1 n=1 n=1

A similar argument holds for the difference.

Remark 1.9. Convergence or divergence of a series is not affected by deleting a finite number
of terms at the beginning of the series.

Example 1.10. Find the sum of the series

∞  
X 3 2
−
k=1
4k 5k − 1

if it exists.

Solution.
∞   ∞ ∞
X 3 2 X 1 X 1 3
− =3 −2 =− .
k=1
4k 5k − 1 k=1
4k k=1
5k − 1 2

1.1 Divergence and Convergence Tests for Series

Theorem 1.11 (Divergence Theorem). If limk→∞ ak ̸= 0 or does not exist, then the series
P ∞
k=1 ak diverges.

Example 1.12. Determine whether the series ∞ k

P
k=1 k+1 converges or diverges.

4 KPIS. Kimuli
 1.1 Divergence and Convergence Tests for Series

k
Solution. Here, ak = k+1
=⇒ limk→∞ ak = 1 ̸= 0. By the divergence test (Theorem 18),
the series diverges.

Remark 1.13. The converse P∞ of Theorem 18 is not true: limn→∞ an = 0 does not guarantee
convergence. For example, n=1 n1 diverges.

Theorem 1.14 (Integral Test). Let ∞

P
n=1 an be a series with positive terms, and let f (x)
be a function Rsuch that f (n) = an . If f (x) is decreasing and continuous for x ≥ 1, then
P ∞ ∞
n=1 an and 1 f (x) dx both converge or both diverge.

Example 1.15. Using the integral test, determine the convergence of the following series:
P∞ 1
(a) n=1 n2
P∞ 1
(b) n=1 n (harmonic series)
P∞ n
(c) n=1 en2

1 1
Solution. (a) an = n2
, f (x) = x2
is decreasing and continuous for x ≥ 1.
Z ∞ ∞
1 X 1
2
dx = 1 < ∞ =⇒ 2
converges.
1 x n=1
n

(b) an = n1 , f (x) = 1
x
is decreasing and continuous for x ≥ 1.
Z ∞ ∞
1 X 1
dx = ∞ =⇒ diverges.
1 x n=1
n

n x
(c) an = en2
, f (x) = ex2
is decreasing and continuous for x ≥ 1.
Z ∞ ∞
x 1 X n
2 dx = < ∞ =⇒ n2
converges.
1 ex 2e n=1
e
P∞ 1
Theorem 1.16 (p-Series Test). For the series n=1 an with an = np
, the series converges
if p > 1 and diverges if 0 < p ≤ 1.

Proof. Use the integral test:

∞ Z ∞
X 1 1
p
converges ⇐⇒ dx converges.
n=1
n 1 xp

P∞
Theorem 1.17 (Root Test). For the series n=1 an , define
p
n
L = lim |an |.
n→∞

5 KPIS. Kimuli
Chapter 1: Infinite Series and Convergence

(a) If L > 1, the series diverges.

(b) If L < 1, the series converges.

(c) If L = 1, the test is inconclusive.

P∞ nn
Example 1.18. Determine the convergence of the series n=1 32n using the root test.
nn
Solution. Here an = 32n
, so
r
n nn n
L = lim n
= lim = ∞.
n→∞ 32 n→∞ 32

Since L > 1, the series diverges by the root test.

Theorem 1.19 (Comparison Test). Suppose ∞
P P∞
n=1 an and n=1 bn are series with an , bn ≥ 0
and an ≤ bn for all n.
P P
(a) If bn diverges, then an also diverges.
P P
(b) If bn converges, then an also converges.
Example 1.20. Determine the convergence of ∞ 1
P
n=1 3n+n using the comparison test.

1 1 1
Solution. Let an = 3n+n = 4n ≤ 3n = bn .
P P 1 1
P 1
Check bn = 3n
=3 n
. The harmonic series diverges, but here we check using the
root test: r
n 1 1
L = lim = < 1.
n→∞ 3n 3
P P
Thus bn converges, and by the comparison test, an also converges.
P∞
Theorem 1.21 (D’Alembert’s Ratio Test). Let n=1 an be a series with non-zero terms
and
an+1
l = lim .
n→∞ an
Then
(a) if l < 1, the series converges,

(b) if l > 1, the series diverges,

(c) if l = 1, the test is inconclusive.

Proof. (a) Suppose l < 1. Choose r ∈ (l, 1) and let ϵ = r − l > 0. Then ∃N0 ∈ N such that

an+1
− l < ϵ ∀n ≥ N0 ,
an
which implies |an+1 | < r|an | for all n ≥ N0 . Hence,

|aN0 +1 | ≤ r|aN0 |, |aN0 +2 | ≤ r|aN0 +1 | ≤ r2 |aN0 |, . . . , |aN0 +m | ≤ rm |aN0 |.

6 KPIS. Kimuli
 1.1 Divergence and Convergence Tests for Series

Consider the series

∞
X N
X 0 −1 ∞
X N
X0 −1 ∞
X
|an | = |an | + |an | ≤ |an | + rm |aN0 |.
n=1 n=1 n=N0 n=1 m=0
P∞ m
P
The geometric series
P m=0 r converges since |r| < 1. By the comparison test, |an |
converges, hence an converges.
(b) Suppose l > 1. Then for some N0 ∈ N, we have
|an+1 |
> 1 ∀n ≥ N0 ,
|an |
so |an | is increasing and |an | ≥ |aN0 | =
̸ 0. Hence limn→∞ an ̸= 0, and by the divergence test,
the series diverges. P1 P 1
(c) If l = 1, the ratio test is inconclusive. For example, n
diverges and n2
converges,
yet in both cases
an+1
lim = 1.
n→∞ an

Further tests are required in this case.

Example 1.22. Determine whether the series
∞
X 2n
n=1
n!

converges using the ratio test.

2n
Solution. Let an = n!
. Then

an+1 2n+1 /(n + 1)! 2

L = lim = lim = lim = 0.
n→∞ an n→∞ 2n /n! n→∞ n + 1

Since L = 0 < 1, the series converges by the ratio test.

Theorem 1.23 (Alternating Series Test). An alternating series

a1 − a2 + a3 − a4 + · · · + (−1)k ak + . . . , ak > 0

converges if
(a) a1 ≥ a2 ≥ · · · ≥ ak ≥ ak+1 ≥ . . . (monotone decreasing),

(b) limk→∞ ak = 0.
Example 1.24. Determine whether the series
∞
X (−1)n
n=1
n
converges.

7 KPIS. Kimuli
Chapter 1: Infinite Series and Convergence

Solution. Here ak = k1 is monotone decreasing and limk→∞ ak = 0. Hence, by the alternating

series test, the series converges.

Definition 1.25 (Absolute Convergence). A series ∞

P P∞
n=1 a n converges absolutely if n=1 |an |
converges.

Convergence). A series ∞
P
Definition 1.26 (Conditional n=1 an converges conditionally if
P ∞ P ∞
a
n=1 n converges but n=1 |a n | diverges.
P∞ (−1)n
Remark 1.27. The series converges conditionally but not absolutely.
n=1 n
P∞ P∞
Theorem 1.28 (Absolute Convergence Theorem). If n=1 |an | converges, then n=1 an
converges.

Proof. For each n ∈ N,

−|an | ≤ an ≤ |an | =⇒ 0 ≤ an + |an | ≤ 2|an |.

P P P
Since |an | converges, 2|an | converges. By the comparison test, (an + |an |) converges.
Now X X X
an = (an + |an |) − |an | =⇒ (an + |an |) −
an = |an |,
P
which is the difference of two convergent series. Hence an converges.

Example 1.29. Show that the following series converge absolutely:

P∞ (−1)n+1
(a) n=1 n2
P∞
(b) √sin n
n=1 n3 +1

n+1
Solution. (a) (−1) = n12 . Since
P 1
n2 n2
converges by the p-series test (p > 1), the original
series converges absolutely.
(b) √sin n √ 1 √1 = 3/2 1
P 1
3
n +1
≤ 3
n +1
< n3 n
. Since n3/2
converges by the p-series test (p > 1),
√sin n converges absolutely, and hence converges.
P
by the comparison test the series n3 +1

1.2 Special Series

1.2.1 Power Series
Definition 1.30. An infinite series of the form
∞
X
an x n = a0 + a1 x + a2 x 2 + . . .
n=0

is called a power series in x centered at the origin (x = 0).

8 KPIS. Kimuli
 1.2 Special Series

Definition 1.31. An infinite series of the form

∞
X
an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + . . .
n=0

is called a power series in x centered at the point x = c.

Example 1.32. The power series
∞
X
an (x + 3)n
n=0
is centered at x = −3.
Theorem 1.33. For a power series centered at x = c, only one of the following is true:
(a) The series converges only at x = c.
(b) The series converges for all x ∈ R.
(c) There exists a positive number R ∈ R such that the series converges for |x − c| < R
and diverges for |x − c| > R.
Remark 1.34. The real number R is called the radius of convergence and |x − c| < R is
called the interval of convergence.
Definition 1.35. The radius of convergence of a power series centered at x = c is defined
by
an
R = lim , 0 ≤ R ≤ ∞.
n→∞ an+1

Remark 1.36. (a) If R = 0, the series converges only at x = c.

(b) If R = ∞, the series converges for all x.
(c) If R < ∞, the series converges for |x − c| < R, which is the interval of convergence.
Example 1.37. Find the radius and interval of convergence of the following power series:
P∞ xn
(a) n=0 n!
P∞ (−1)n n
(b) n=0 2n (x + 1)
1
Solution. (a) Here an = n!
,c = 0.
an 1/n!
R = lim = lim = lim (n + 1) = ∞.
n→∞ an+1 n→∞ 1/(n + 1)! n→∞

Hence, the series converges

n
for all x.
(b) Here an = (−1)
2n
, c = −1.
an (−1)n /2n
R = lim = lim = lim 2 = 2.
n→∞ an+1 n→∞ (−1)n+1 /2n+1 n→∞

Hence, the series converges for |x + 1| < 2.

9 KPIS. Kimuli
Chapter 1: Infinite Series and Convergence

Definition 1.38 (Taylor Series). If f (x) has derivatives of all orders at x = c, then the
power series
∞
X f (n) (c) f ′′ (c)
f (x) = (x − c)n = f (c) + f ′ (c)(x − c) + (x − c)2 + . . .
n=0
n! 2!

is called the Taylor series of f centered at x = c.

Remark 1.39. If c = 0, the Taylor series is called the Maclaurin series.

Example 1.40 (Taylor Series of f (x) = x1 centered at x = 1). For f (x) = x1 , the derivatives
are:
1 2 6 24
f ′ (x) = − 2 , f ′′ (x) = 3 , f ′′′ (x) = − 4 , f (4) (x) = 5 , . . .
x x x x
The Taylor series at x = 1 is
∞
X f (n) (1)
f (x) = (x − 1)n = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + (x − 1)4 − . . .
n=0
n!

Applications of Series
Power series (including Taylor series) have many applications:

ˆ In Ordinary Differential Equations (ODEs): Series solutions are used for both
ordinary and singular points (regular or irregular singularities).

ˆ In Partial Differential Equations (PDEs): Fourier series and other series meth-
ods represent unknown functions, especially when closed-form solutions are hard or
impossible to find.

Tutorial Sheet III

(a) Determine whether the following series converge, and if they converge, find their sum:
1 1 1
(a) 2·5
+ 5·8 + 8·11 + ...
1 1 1
(b) 1·2
+ 2·3 + 3·4 + ...
1 1 1
(c) 1·2·3
+ 2·3·4 + 3·4·5 + ...
P∞ 1
(d) n=0 n!

(b) Prove that

∞
X n
converges to 1.
n=1
(n + 1)!

10 KPIS. Kimuli
 1.2 Special Series

(c) Determine whether the series

∞
X n−1
n=2
(n + 1)(n + 2)(n + 3)

converges. If it converges, find its sum.

(d) Using the integral test, show that

∞
X 1
, p>0
n=2
n(log n)p

converges for p > 1 and diverges for p ≤ 1.

(e) Using the limit comparison test, determine whether the following series converge:
P∞ n
(a) n=1 n4 −3
P∞ n2
(b) n=1 n3 −3
P∞ n2
(c) n=1 en2
P∞ 3n+1
(d) n=1 4n−1

(f) Using Cauchy’s root test, determine whether the following series converge:
P∞ 1
(a) n=1 3n n
P∞ 1
(b) n=1 22n +(−1)n
P∞ 1
(c) n=1 n3 +(−1)n
P∞ 2 an
(d) n=1 n e

(g) Using the ratio test, determine whether the following series converge:
P∞ 10n
(a) n=1 n
P∞ an
(b) n=1 n! , a > 0
P∞ log n
(c) n=1 2n
P∞ n2
(d) n=1 n3 +1

(h) Prove or give a counterexample: If

∞
X ∞
X ∞
X
an and bn converge, then an bn converges.
n=1 n=1 n=1

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