Convergence Tests
Academic Resource Center
Series
• Given a sequence {a0, a1, a2,…, an }
• The sum of the series, Sn=
n
a
k 1
k
• A series is convergent if, as n gets larger and larger, Sn goes to
some finite number.
• If Sn does not converge, and Sngoes to ∞, then the series is
said to be divergent
Geometric and P-Series
• The two series that are the easiest to test are geometric series
and p-series.
• Geometric is generally in the form

 ar
• P-series is generally in the form
k 1




n 1
1
n
p
k
Geometric Series
• A geometric series is a series in which there is a constant ratio
between successive terms
• 1 +2 + 4 + 8 + … each successive term is the previous term
multiplied by 2
•
each successive term is the previous term
squared.
1
2


1
4

1
16

1
256
 ...
Geometric Series
• Sn =

 =arar +ar^2 + ar^3 + … +ar^k
k
k 1
• As a result, if |r|<1, the geometric series will converge to
, and if |r| 1 the series will diverge.


a
1 r


P-Series
• Given a series


n 1
1
n
p
• This series is said to be convergent if p>1,

• And divergent if p
1


Geometric and P-Series
Examples

3

4
n 1

n 1
So S= 3/(1-3/4) = 12




n 1
The Sum of the series, S
S 
a
1 r

3

1
 1 2
   
 n 
n
n 1
1
This series is
geometric with a=3
and r =3/4. Since
r<1, this series will
converge.
Here, p=3, so p>1. Therefore our series
will converge
1
n
n 1
n 1
 3 
  3 
 4 
n 1
n
Here, p=1/2, so p<1. Therefore our series will
diverge
Convergence Tests
•
•
•
•
•
•
•
Divergence test
Comparison Test
Limit Comparison Test
Ratio Test
Root Test
Integral Test
Alternating Series Test

Divergence Test
•
•
•
•
•

Say you have some series
 an
The easiest way to see if a series diverges
n  0 is this test
Evaluate L= Lim
If L 0, the series diverges
If L=0, then this test is inconclusive
a 
n 



n

Divergence Test Example

 5n
n
2
2
4
n 1
Lim

n 
n
n 1
2
2
5n  4


Let’s look at the limit
of the series
 Lim
n 
n
2
5n
2

1
5
 0
Therefore, this series is divergent
1
n
Lim
n 
2
1
n
2
 0
The limit here is equal to zero, so this test is inconclusive.
However, we should see that
this a p-series with p>1, therefore this will converge.

Comparison Test
• Often easiest to compare geometric and p series.
• Let
• If
 ak
• If
• If

and
be series with non-negative terms.
for
b as k gets big, then
 k,
k
converges, then
diverges, then
b k a k
b
k
b
k
a
a


converges
diverges


k
k
Comparison Test Example

3
1
n
n 1
Test to see if this series converges
using the comparison test
1

This is very similar to
1
3
which is a geometric series so
it will converge
n
n 1


And since
3
 n 1

1
n


3
n 1
1
n
1
our original series will also converge
Limit Comparison Test
• Let
and b
be series with non-negative terms.
 ak
 k
• Evaluate Lim
• If lim=L, some finite number,
then both
and
ak
either converge or diverge.


bk
k 
•
and
are generally
geometric series or p-series, so
 ak
seeing whether
these
series
are
convergent
is
fast.

b
k
 ak




b
k


Limit Comparison Test
Example

9
n
 3  10
Determine whether
this series
converges or not
n
n 1

Compare it with

n 1

9
n
10
n
n

9
n
 9 
   
10 
n 1
n
n
so
Lim 3  10n  1  0
n    9 
 
10 
 9 
And since    Is a geometric series with r<1, this series converges,
10  therefore so does our original series
n 1



Ratio Test
• Let
a
be a series with non-negative terms.
k
• Evaluate L= Lim

a k 1
• If L < 1, then
converges
ak
kdiverges

• If L > 1, then
 isa kinconclusive
• If L = 1, then this test




a
k
Ratio Test Example
Test for convergence

  1
n
n 1
(  1)
n 1
( n  1)
3
3
an
3
3
Lim (
n 
n
n 1
n
n
3
( n  1)
 Lim
 3
n 1
n 
3
n
3
n
3
(  1) n
n 

3
n
a n 1
Look at the limit of
n 1
Lim
1
n
3
3
) 
1
3
Lim (1 
n 
1
n
3
) 
1
3
1
Since L<1, this series
will converge based
on the ratio test
Root Test
• Let
be a series with non-negative terms.
 an
• Useful if involves nth powers
a
• Evaluate L= Limn
1

• If L < 1,
is convergent
(a n ) n
n 
• If L > 1,
is divergent
an
•If L = 1, thenthis
test is inconclusive



a

n
Root Test Example
Test for convergence

4n  5
 ( 5n  6 )
1
Lets evaluate the limit, L =Lim (a n ) n
n
n 
n 1
Lim ((
n 

4n  5
5n  6

1
n
) )
n
 Lim
n 
4n  5
5n  6
By the root test, since L<1, our series will converge.

4
5
1
Integral Test
•
•
•
•
•
Given the series
, let  =a f(k)
ak
k
f must be continuous, positive, and decreasing for x > 0
will converge only if
converges.
If
diverges,
 then the series will also diverge.



In
general,
however,
 ak
f ( x ) dx

k0
0


f ( x ) dx
0





k 1

ak 


1
f ( x ) dx
Integral Test Example
Test for convergence

1
 (2 n  1)
f (x) 
So let
3
Since x>0, f(x) is
continuous and
1
positive.
3
is negative so
(2 x  1) f’(x)
we know f(x) is
decreasing.
n 1
Now let’s look
at the integral



1
1
(2 x  1)
  Lim (
t 
3
dx  
1
(2 t  1)
2

1
2
 2 Lim [
t 
1
( 3)
2
)
1
(2 x  1)
2
]
t
1
1
9
Since the integral converged to a finite number, our original series will also converge
Note: Series
 will most likely not converge to 1/9, but it will converge nonetheless.
Alternating Series Test
• Given a series
• IF
•
• Lim
 (1)
k
for all k, and
= 0
a k 1  a k
Then the series is convergent
k 



ak
ak

, where ais positive for all k
k


Alternating Series Test
Example
Test for convergence

  1
n 
n
Check:
Is this series decrease- yes
Is the Lim=0?
3
2
0
3
n 4

Therefore,

2
n 4
n 1
Lim
n
n 1
  1
n 1
Yes
n 1
n
3
2
n 4
, is convergent.
More Examples

1.

n 1


3.

cos n 
n
2.
3 4
n 1
10
n
 ( n  1) 4

2 n 1
n 1




4.
1 n  n
2
1 n  n
1
 n ln n
n2
2
6
Answers
•
•
•
•
1. By Alternating series test, series will converge
2. By the comparison test, series will diverge
3. By the ratio test, series will converge
4. By the integral test, series will diverge