SECTION 8.4
OTHER CONVERGENCE
TESTS
INFINITE SEQUENCES AND SERIES
The convergence tests that we have looked at so
far apply only to series with positive terms.
In this section we learn how to deal with series
whose terms are not necessarily positive.
8.4 P2
ALTERNATING SERIES
An alternating series is a series whose terms
are alternately positive and negative.

Here are two examples:

1 1 1 1 1
(1) n 1
1       ...  
2 3 4 5 6
n
n 1

1 2 3 4 5 6
n
n
       ...   (1)
2 3 4 5 6 7
n 1
n 1
8.4 P3
ALTERNATING SERIES
From these examples, we see that the nth term
of an alternating series is of the form
an = (–1)n – 1bn or an = (–1)nbn
where bn is a positive number.

In fact, bn = |an|
8.4 P4
ALTERNATING SERIES
The following test states that, if the terms of an
alternating series decrease toward 0 in absolute
value, the series converges.
8.4 P5
ALTERNATING SERIES TEST
If the alternating series

 (1)
n 1
n 1
bn  b1  b2  b3  b4  b5  b6  ...
(bn  0)
satisfies
(i) bn+1 ≤ bn for all n
(ii) lim bn  0
n 
then the series is convergent.
8.4 P6
ALTERNATING SERIES
Before giving the proof, let’s look Figure 1,
which gives a picture of the idea behind the
proof.
8.4 P7
ALTERNATING SERIES
First, we plot s1 = b1 on a number line.
To find s2,we subtract b2.

So, s2 is to the left of s1.
8.4 P8
ALTERNATING SERIES
Then, to find s3, we add b3.

So, s3 is to the right of s2.
However, since b3 < b2, s3 is to the left of s1.
8.4 P9
ALTERNATING SERIES
Continuing in this manner, we see that the
partial sums oscillate back and forth.

Since bn → 0, the successive steps are becoming
smaller and smaller.
8.4 P10
ALTERNATING SERIES
The even partial sums s2, s4, s6, … are
increasing.
The odd partial sums s1, s3, s5, … are decreasing.
8.4 P11
ALTERNATING SERIES
Thus, it seems plausible that both are
converging to some number s, which is the sum
of the series.

Therefore, in the following proof we consider the
even and odd partial sums separately.
8.4 P12
PROOF OF THE ALTERNATING SERIES
TEST
First, we consider the even partial sums:
s2 = b 1 – b 2 ≥ 0
since b2 ≤ b1
s4 = s2 + (b3 – b4) ≥ s2
since b4 ≤ b3
In general,
s2n = s2n – 2 + (b2n – 1 – b2n) ≥ s2n – 2
since b2n ≤ b2n – 1
Thus, 0 ≤ s2 ≤ s4 ≤ s6 ≤ … ≤ s2n ≤ …
8.4 P13
PROOF OF THE ALTERNATING SERIES
TEST
However, we can also write:
s2n = b1 – (b2 – b3) – (b4 – b5) – …
– (b2n – 2 – b2n – 1) – b2n


Every term in brackets is positive.
So, s2n ≤ b1 for all n.
Thus, the sequence {s2n} of even partial sums is
increasing and bounded above.

Therefore, it is convergent by the Monotonic
Sequence Theorem.
8.4 P14
PROOF OF THE ALTERNATING SERIES
TEST
Let’s call its limit s, that is,
lim s2 n  s
n 

Now, we compute the limit of the odd partial sums:
lim s2 n 1  lim( s2 n  b2 n 1 )
n 
n 
 lim s2 n  lim b2 n 1
n 
 s0
s
n 
(condition ii)
8.4 P15
PROOF OF THE ALTERNATING SERIES
TEST
As both the even and odd partial sums converge
to s, we have
lim sn  s
n 

See Exercise 46 in Section 8.1
Thus, the series is convergent.
8.4 P16
Example 1
The alternating harmonic series

1 1 1
(1) n 1
1     ...  
2 3 4
n
n 1
satisfies
1
1

(i) bn+1 < bn
because
n 1 n
1
bn  lim  0
(ii) lim
n 
n  n

It is convergent by the Alternating Series Test.
8.4 P17
ALTERNATING SERIES
Figure 2 illustrates
Example 1 by showing
the graphs of the terms
an = (–1)n – 1/n and the
partial sums sn.
8.4 P18
ALTERNATING SERIES
Notice how the values
of sn zigzag across the
limiting value, which
appears to be about 0.7

In fact, it can be proved
that the exact sum of the
series is ln 2 ≈ 0.693.
8.4 P19
Example 2
(1) 3n
The series 
is alternating.
n 1 4n  1

n
However,
3n
3
3
lim bn  lim
 lim

n 
n  4n  1
n 
1 4
4
n

So, condition (ii) is not satisfied.
8.4 P20
Example 2 SOLUTION
Instead, we look at the limit of the nth term
of the series:
n
(1) 3n
lim an  lim
n 
n  4 n  1
This limit does not exist.

So, the series diverges by the Test for Divergence.
8.4 P21
Example 3
Test the series

2
n
n 1
(

1)
 3

n 1
n 1
for convergence or divergence.
SOLUTION


The given series is alternating.
So, we try to verify conditions (i) and (ii) of the
Alternating Series Test.
8.4 P22
Example 3 SOLUTION
Unlike the situation in Example 1, it is not
obvious that the sequence given by bn = n2/(n3 +
1) is decreasing.
However, if we consider the related function f(x)
= x2/(x3 + 1), we find that:
3
x(2  x )
f '( x)  3
2
( x  1)
8.4 P23
Example 3 SOLUTION
Since we are considering only positive x, we see
that f ’(x) < 0 if 2 – x3 < 0, that is, x > 3 2 .
Thus, f is decreasing on the interval ( 3 2 , ∞).
This means that f(n + 1) < f(n) and, therefore,
bn+1 < bn when n ≥ 2.


The inequality b2 < b1 can be verified directly.
However, all that really matters is that the sequence
{bn} is eventually decreasing.
8.4 P24
Example 3 SOLUTION
Condition ii is readily verified:
1
n
n2
lim bn  lim 3
 lim
0
n 
n  n  1
n 
1
1 3
n

Thus, the given series is convergent by the
Alternating Series Test.
8.4 P25
ESTIMATING SUMS
A partial sum sn of any convergent series can be
used as an approximation to the total sum s.


However, this is not of much use unless we can
estimate the accuracy of the approximation.
The error involved in using s ≈ sn is the remainder
Rn = s – sn.
8.4 P26
ESTIMATING SUMS
The next theorem says that, for series that
satisfy the conditions of the Alternating Series
Test, the size of the error is smaller than bn+1.

This is the absolute value of the first neglected term.
8.4 P27
ALTERNATING SERIES ESTIMATION
THEOREM
If s = Σ(–1)n–1bn is the sum of an alternating
series that satisfies
(i) 0 ≤ bn+1 ≤ bn
and
(ii) lim bn  0
n 
then
|Rn| = |s – sn| ≤ bn+1
8.4 P28
ALTERNATING SERIES ESTIMATION
THEOREM PROOF
From the proof of the Alternating Series Test,
we know that s lies between any two
consecutive partial sums sn and sn+1.
It follows that:
|s – sn| ≤ |sn+1 – sn| = bn+1
8.4 P29
ALTERNATING SERIES ESTIMATION
THEOREM
You can see geometrically why the theorem is
true by looking at Figure 1.


Notice that s – s4 < b5, |s – s5| < b6, and so on.
Notice also that s lies between any two consecutive
partial sums.
8.4 P30
Example 4
Find the sum of the series

(1) n

n!
n 0
correct to three decimal places.

By definition, 0! = 1.
8.4 P31
Example 4 SOLUTION
First, we observe that the series is convergent
by the Alternating Series Test because:
1
1
1

  bn
(i)
(n  1)! n !( n  1) n !
1 1
1
(ii) 0    0 so bn   0 as n  
n! n
n!
8.4 P32
Example 4 SOLUTION
To get a feel for how many terms we need to
use in our approximation, let’s write out the first
few terms of the series:
1 1 1 1 1 1 1 1
s          ...
0! 1! 2! 3! 4! 5! 6! 7!
1
1
1
 1  1  12  16  241  120
 720
 5040
 ...
8.4 P33
Example 4 SOLUTION

Notice that
1
1
b7  5040
 5000
 0.0002
and
1
1
s6  1  1  12  16  241  120
 720
 0.368056
8.4 P34
Example 4 SOLUTION
By the Alternating Series Estimation Theorem,
we know that:
| s – s6 | ≤ b7 < 0.0002


This error of less than 0.0002 does not affect the
third decimal place.
So, we have s ≈ 0.368 correct to three decimal
places.
8.4 P35
NOTE
The rule that the error (in using sn to
approximate s) is smaller than the first neglected
term is, in general, valid only for alternating
series that satisfy the conditions of the
Alternating Series Estimation Theorem.

The rule does not apply to other types of series.
8.4 P36
ABSOLUTE CONVERGENCE
Given any series Σan, we can consider the
corresponding series

a
n 1
n
 a1  a2  a3  ...
whose terms are the absolute values of the terms
of the original series.
8.4 P37
Definition
A series Σan is called absolutely convergent if the
series of absolute values Σ|an| is convergent.
Notice that if Σan is a series with positive terms,
then |an| = an and so absolute convergence is the
same as convergence.
8.4 P38
Example 5
The series
n 1
(1)
1 1 1
 1  2  2  2  ...

2
n
2 3 4
n 1
is absolutely convergent because



n 1

(1) n 1
1
1 1 1
  2  1  2  2  2  ...
2
n
2 3 4
n 1 n
is a convergent p-series (p = 2).
8.4 P39
Example 6
We know that the alternating harmonic series
(1)

n
n 1

n 1
1 1 1
1     ...
2 3 4
is convergent.

See Example 1.
8.4 P40
Example 6
However, it is not absolutely convergent
because the corresponding series of absolute
values is:

(1) n 1  1
1 1 1
 1     ...

n
2 3 4
n 1
n 1 n

This is the harmonic series (p-series with p = 1) and
is, therefore, divergent.
8.4 P41
Definition
A series Σan is called conditionally convergent
if it is convergent but not absolutely convergent.
Example 6 shows that the alternating harmonic
series is conditionally convergent.


Thus, it is possible for a series to be convergent but
not absolutely convergent.
However, the next theorem shows that absolute
convergence implies convergence.
8.4 P42
Theorem 1
If a series Σan is absolutely convergent, then it is
convergent.
PROOF

Observe that the inequality
0  an  an  2 an

is true because |an| is either an or –an.
8.4 P43
Theorem 1 PROOF
If Σan is absolutely convergent, then Σ|an|
is convergent.
So, Σ2|an| is convergent.

Thus, by the Comparison Test, Σ(an + |an|) is
convergent.
Then,
 a  ( a
n
n
 an )   an
is the difference of two convergent series and is,
therefore, convergent.
8.4 P44
Example 7
Determine whether the series

cos n cos1 cos 2 cos 3
 2  2  2  ...

2
1
2
3
n 1 n
is convergent or divergent.
8.4 P45
Example 7 SOLUTION

cos n cos1 cos 2 cos 3
 2  2  2  ...

2
1
2
3
n 1 n
The series has both positive and negative terms,
but it is not alternating.



The first term is positive.
The next three are negative.
The following three are positive—the signs change
irregularly.
8.4 P46
Example 7 SOLUTION
We can apply the Comparison Test to the series
of absolute values:


cos n
cos n
 2

2
n
n
n 1
n 1
8.4 P47
Example 7 SOLUTION
Since |cos n| ≤ 1 for all n, we have:
cos n
1
 2
2
n
n


We know that Σ1/n2 is convergent (p-series with p =
2).
Hence, Σ(cos n)/n2 is convergent by the Comparison
Test.
Thus, the given series Σ(cos n)/n2 is absolutely
convergent and, therefore, convergent by
Theorem 1.
8.4 P48
THE RATIO TEST
The following test is very useful in determining
whether a given series is absolutely convergent.
8.4 P49
THE RATIO TEST
i)
If lim an1  L  1 , then the series
n 
an

a
n 1
n
is
absolutely convergent (and therefore
convergent).
ii) If lim an1  L  1 or lim an1   ,
n 
an

n 
an
then the series  an is divergent.
n 1
iii) If lim an1  1 , the Ratio Test is inconclusive;
n 
an
that is, no conclusion can be drawn about
the convergence or divergence of Σan.
8.4 P50
THE RATIO TEST (i) PROOF
The idea is to compare the given series with a
convergent geometric series.

Since L < 1, we can choose a number r such that L <
r < 1.
Since
an 1
lim
L
and L  r
n  a
n
the ratio |an+1/an| will eventually be less than r.

That is, there exists an integer N such that:
an 1
r
whenever n  N
an
8.4 P51
THE RATIO TEST (i) PROOF
Equivalently,
|an+1| < |an|r
whenever n ≥ N
Putting n successively equal to N, N + 1, N +
2, … in Equation 4, we obtain:
|aN+1| < |aN|r
|aN+2| < |aN+1|r < |aN|r2
|aN+3| < |aN+2|r < |aN|r3
In general,
|aN+k| < |aN|rk for all k ≥ 1
8.4 P52
THE RATIO TEST (i) PROOF
Now, the series

k
2
3
a
r

a
r

a
r

a
r
 ...
 N
N
N
N
k 1
is convergent because it is a geometric series
with 0 < r < 1.
Thus, the inequality 3, together with the
Comparison Test, shows that the series


n  N 1

an   aN  k  aN 1  aN  2  aN 3  ...
k 1
is also convergent.
8.4 P53
THE RATIO TEST (i) PROOF

It follows that the series
a
n 1
n
is convergent.
Recall that a finite number of terms doesn’t
affect convergence.

Therefore, Σan is absolutely convergent.
8.4 P54
THE RATIO TEST (ii) PROOF
If |an+1/an| → L > 1 or |an+1/an| → ∞, then the
ratio |an+1/an| will eventually be greater than 1.

That is, there exists an integer N such that:
an 1
 1 whenever n  N
an
This means that |an+1| > |an| whenever n ≥ N, and
so
lim an  0
n 

Therefore, Σan diverges by the Test for Divergence.
8.4 P55
NOTE
Part (iii) of the Ratio Test says that, if
lim an 1 / an  1
n 
the test gives no information.
For instance, for the convergent series Σ1/n2, we
have:
an 1
an
1
2
2
n
1
(n  1)



1
2
2
1
(n  1)
 1
1

2


n
 n
as n  
8.4 P56
NOTE
an 1 / an  1 , the series Σan
Therefore, if lim
n 
might converge or it might diverge.


In this case, the Ratio Test fails.
We must use some other test.
8.4 P57
Example 8

3
n
n
Test the series  (1) n
3
n 1
for absolute convergence.
SOLUTION

We use the Ratio Test with an = (–1)n n3 / 3n, as
follows.
8.4 P58
Example 8 SOLUTION
an 1
an
(1) n 1 (n  1)3
3
n
n 1
(n  1) 3
3

 n 1  3
n 3
(1) n
3
n
n
3
1  n 1  1  1 
1
 
  1     1
3 n  3 n 
3
3
Thus, by the Ratio Test, the given series is
absolutely convergent and, therefore,
convergent.
8.4 P59
Example 9

nn
Test the convergence of the series 
n 1 n !
SOLUTION

Since the terms an = nn/n! are positive, we don’t
need the absolute value signs.
8.4 P60
Example 9 SOLUTION
an 1 (n  1) n 1 n ! (n  1)(n  1) n n !

 n 
 n
an
(n  1)! n
(n  1)n !
n
 n 1  1 

  1    e
 n   n
n

n
as n  
Since e > 1, the series is divergent by the Ratio Test.
8.4 P61
ABSOLUTE CONVERGENCE
The following test is convenient to apply when
nth powers occur.

Its proof is similar to the proof of the Ratio Test
and is left as Exercise 43.
8.4 P62
THE ROOT TEST
i)
If lim n an  L  1, then the series
n 

is
a
 n
n 1
absolutely convergent (and therefore
convergent).
n a  L  1 or lim n a  
ii) If lim
, then the
n
n
n 
n 

series  an is divergent.
n 1
n a 1
lim
iii) If n n
, the Root Test is inconclusive.
8.4 P63