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Convergence Tests for Infinite Series

In this tutorial, we review some of the most common tests for the convergence of an infinite series

¥
å
 k = 0 
 ak = a0 + a1 + a2 + ¼
The proofs or these tests are interesting, so we urge you to look them up in your calculus text.

Let

s0
=
a0
s1
=
a1
:
sn
=
n
å
 k = 0 
 ak
:
If the sequence { sn } of partial sums converges to a limit L, then the series is said to converge to the sum L and we write

                            
¥
å
 k = 0 
 ak = L.
                 
For j ³ 0, ¥
å
k = 0		ak converges if
and only if¥
å
k = j 		ak converges,
so in discussing convergence
we often just write åak.

Example

Consider the geometric series

¥
å
 k = 0 
 xk.
The nth partial sum is
sn = 1 + x + x2 + ¼+ xn.
Multiplying both sides by x,
xsn = x + x2 + x3 + ¼+ xn+1.
Subtracting the second equation from the first,
(1-x)sn = 1-xn+1,
so for x ¹ 1,
sn = 1-xn+1
1-x
 .
For |x| < 1,

lim
n ® ¥ 
 sn = 1
1-x
 .

It is easy to see that ¥
å
k = 0		 xk diverges for |x| ³ 1.
Thus ¥
å
k = 0		 xk = 1/(1-x) for |x| < 1 and diverges for |x| ³ 1.

Divergence Test

If  
lim
k ® ¥		 ak ¹ 0, then ¥
å
k = 0		 ak diverges.

Example

The series ¥
å
k = 0		 k/(2k+1) diverges, since  
lim
k ® ¥		 k/(2k+1) = 1/2 ¹ 0.

Integral Test

Let f(x) be continuous, decreasing, and positive for x ³ 1.
Then ¥
å
k = 1		 f(k) converges if and only if ò ¥
1 		f(x)dx converges.

Example

Consider the p-series
¥
å
 k = 1 
 1
kp
 = 1
1p
 + 1
2p
 + 1
3p
 + ¼
Since
ó
õ 		¥

1 
 1
xp
 dx = ì
ï
ï
ï
ï
ï
í
ï
ï
ï
ï
ï
î
1
1-p
 x1-p ê
ê
ê 		¥

1 
,    p > 1
ln|x| |1¥
,    p = 1
1
1-p
 x1-p ê
ê
ê 		¥

1 
,    0 < p < 1
 = ì
ï
ï
ï
í
ï
ï
ï
î
1
1-p
¥
¥,
the series converges for p > 1 and diverges for 0 < p £ 1.
The divergent p-series
¥
å
 k = 1 
 1
k
with p = 1 is called the Harmonic Series.

Comparison Test

Let åak and åbk be series with non-negative terms. If ak £ bk for all k sufficiently large, then

  1. If åbk converges, then åak also converges.

  2. If åak diverges, then åbk also diverges.

Informally, if the ``larger'' series converges, so does the ``smaller.'' If the ``smaller'' series diverges, so does the ``larger.''

Examples

  • Since ¥
    å
    k = 1    		 1/(k2) converges,
    so does ¥
    å
    k = 1    		 1/(k2 + 3).
    1/(k2 +3) < 1/(k2) for all k

  • Since ¥
    å
    k = 1    		 1/k diverges,
    so does ¥
    å
    k = 1    		 1/(ln|k+1|).
    1/(ln|k+1|) > 1/k for k ³ 2

Limit Comparison Test

Let åak and åbk be series with positive terms. If

lim
k ® ¥ 
 ak
bk
 = L
where 0 < L < ¥ then åak and åbk either both converge or both diverge.

Example

The series ¥
å
k = 1		 (k2-1)/( 5k3) diverges, since ¥
å
k = 1		 1/k diverges and


lim
k ® ¥ 
k2-1
5k3
1
k
=
lim
k ® ¥ 
 k2-1
5k2
 = 1
5
 .

Ratio Test

Let åak be a series with positive terms and suppose that

lim
k ® ¥ 
 ak+1
ak
 = L.

  1. If L < 1, then åak converges.

  2. If L > 1, then åak diverges.

  3. If L = 1, then the test is inconclusive.

Example

The series ¥
å
k = 1		 1/k! converges, since


lim
k ® ¥ 
1
(k+1)!
1
k!
=
lim
k ® ¥ 
 1
k+1
 = 0.

Root Test

Let åak be a series with non-negative terms and suppose that

lim
k ® ¥ 
 (ak)1/k = L.

  1. If L < 1, then åak converges.

  2. If L > 1, then åak diverges.

  3. If L = 1, then the test is inconclusive.

Example

The series ¥
å
k = 0		 [k/( 2k+1)]k converges, since


lim
k ® ¥ 
 é
ê
ë 		æ
ç
è 		k
2k+1
 ö
÷
ø 		k

 
 ù
ú
û 		1/k

 
 =
lim
k® ¥ 
 k
2k+1
 = 1
2
 .

Alternating Series Test

Consider the alternating series
¥
å
 k = 0 
 (-1)kak
where ak > 0 for all k ³ 0.

If ak+1 < ak for all k and lim ak = 0, then ¥
å
k = 0		 (-1)kak converges.

Example

The series

¥
å
k = 0		 (-1)k
(k+1)
converges, since 1/((k+1) + 1) < 1/(k+1) and  
lim
k ® ¥		 1/(k+1) = 0.
This series is conditionally convergent, rather than absolutely convergent, since

¥
å
k = 0		 | ((-1)k)/(k+1) | = ¥
å
k = 0		 1/(k+1)

diverges.

Key Concepts

The infinite series

¥
å
 k = 0 
 ak
converges if the sequence of partial sums converges and diverges otherwise.

For a particular series, one or more of the common convergence tests may be most convenient to apply.