LESSON 21 ALTERNATING SERIES
Definition A series whose terms are alternately positive and negative is called an
alternating series.

Theorem (The Alternating Series Test) Let
1)
lim a n  0 and 2) the sequence
n  
a
nN
n
be an alternating series. If
 a  is decreasing, that is
n
an
 1
 an
for all n  N , then the alternating series is convergent.
Proof Will be provided later.
NOTE: The Alternating Series Test (AST) can only give you an answer of
convergence for an alternating series. The divergence of an alternating series
would have to be established by the Divergence Test.
Examples Determine whether the following alternating series converge or diverge
if possible.

1.
 (  1)
n  1
n 1
n
n 4
2
n
Let a n  (  1 )
 1
n
n 4
2
a n = lim
Then nlim
n  
 
Let f ( x ) 
1
n
n
0
0
= nlim
=
 
4
n  4
1

0
1 2
n
2
x
x2  4
We want to show that f is decreasing on an interval of the form [ N ,  ) ,
where N is a positive integer.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
x 2  4  x(2 x)
x2  4  2x2
4  x2
f ( x ) 
=
=
( x 2  4) 2
( x 2  4) 2
( x 2  4) 2
f ( x )  0  x   2
f  is defined for all real numbers

Sign of f ( x ) :

2
Since f ( x )  0 for all x in the open interval ( 2 ,  ) , then the function f
x
is decreasing on the closed interval [ 2 ,  ) . Since f ( x )  2
, then
x 4
n
f (n)  2
 a n . Thus, the sequence
an
is decreasing for all
n  4
n  2.


 (  1)
Thus, the series
n  1
n 1

n
converges by the Alternating Series Test
n 4
2
(AST).
Answer: Converges; AST

2.

n2
(  1) n
3n  4
5n  7
n
Let a n  (  1 )
3n  4
5n  7
Copyrighted by James D. Anderson, The University of Toledo
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4
3n  4
30 3
n
lim
 .
= nlim
=
=
n  
7
  5n  7
5

0
5
5
n
3
an
Then nlim
 
a n does not exist. Thus, lim a n  0 .
Then nlim
 
n  


Thus, the series
(  1) n
n2
3n  4
diverges by the Divergence Test (Div Test)
5n  7
Answer: Diverges; Div Test

3.

(  1) n
 1
ln n
n2
3
n2
n
Let a n  (  1)
 1
ln n
3
n2
a n , we will use L’Hopital’s Rule.
In order to calculate nlim
 
lim
ln x
x   3
x2
Thus, nlim
 
Let f ( x ) 
1
x
3
1
3
lim
(0)  0
= xlim
=
=
  2
 1/ 3
x   x 2/3
2
2
x
3
ln n
n2
3
ln x
3
x2
We want to show that f is decreasing on an interval of the form [ N ,  ) ,
where N is a positive integer.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1 2/3
2
2
x
 ( ln x ) x  1/ 3
x  1 / 3  ( ln x ) x  1 / 3
3
3
f ( x )  x
=
=
4/3
4/3
x
x
1  1/ 3
x ( 3  2 ln x )
3  2 ln x
3
=
3 x 5/ 3
x 4/3
f ( x )  0  3  2 ln x  0  ln x 
3
 x  e 3/ 2
2
f  is defined for all real numbers x such that x  2

Sign of f ( x ) :

2

e 3/ 2
3/ 2
Since f ( x )  0 for all x in the open interval ( e ,  ) and
e 3/ 2 
e 3  e e  4.5 , then the function f is decreasing on the closed
ln x
ln n
f
(
x
)

f
(
n
)

 a n . Thus, the
[
5
,

)
interval
. Since
2
2 , then
3
3
n
x
sequence
 a  is decreasing for all
n

Thus, the series

(  1) n
 1
ln n
3
n2
n2
n  5.
converges by the Alternating Series Test
(AST).
Answer: Converges; AST

4.
 (  1)
n 1
n  1
1
n
Copyrighted by James D. Anderson, The University of Toledo
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n
Let a n  (  1 )
 1
1
n
a n = lim
Then nlim
n  
 
1
0
n
Since n  n  1 for all n  1 , then
Thus, a n  a n
for all n  1 .
 1
for all n  1 . Thus, the sequence

Thus, the series
1
1

for all n  1 .
n n 1
 (  1)
n  1
n 1
 a  is decreasing
n
1
converges by the Alternating Series Test
n
(AST).
Answer: Converges; AST

5.
 (  1)
n0
n
1
( 2 n )!
Recall that 0!  1
n
Let a n  (  1 )
an 
1
( 2 n )!
1 1 1
1
1
1
1








=
1 2 3
2n  2 2n  1 2n
( 2 n )!
NOTE: There are 2 n terms from 1 to 2 n . Thus, there are 2 n fractions
being multiplied. The first fraction is 1. The second fraction is one-half. All
the fractions after the fraction of one-half are less than one-half. Thus, we
obtain that
Copyrighted by James D. Anderson, The University of Toledo
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1
1 1 1
1
1
1
1


 
an 
=   
1 2 3
2n  2 2n  1 2n  2 
( 2 n )!
n  1.
1
Since  
2
n  1.
2n  1
2n
1 1
=    
2 2
1
2n  1
for all
n
n
1
1
= 2   , then 0  a n  2   for all
4
4
n
n
1
1
0  0 and lim 2   = 2 lim    2 ( 0 )  0 , then
Since nlim
n  
n  
 
4
4
1
 0 by the Sandwich Theorem.
lim a n = lim
n   ( 2 n )!
n  
Since [ 2 ( n  1) ] ! = ( 2 n  2 ) ! = ( 2 n  2 ) ( 2 n  1) ( 2 n ) !  ( 2 n ) ! for all
1
1

n  0 , then
for all n  0 .
[ 2 ( n  1) ] ! ( 2 n ) !
Thus, a n  a n
for all n  0 .
 1
for all n  0 . Thus, the sequence

Thus, the series
 (  1)
n
n0
 a  is decreasing
n
1
converges by the Alternating Series Test
(2n) !
(AST).
Answer: Converges; AST

x 2n
NOTE: We will show in a later lesson that cos x   (  1) ( 2 n )! . Thus,
n0
n

 (  1)
n0
n
1
 cos 1 .
( 2 n )!
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

Theorem If
a
nN
n
is a convergent alternating series, then the error involved in
approximating the sum S of the series by the n th partial sum
Sn  aN  aN 1  aN  2       aN  n 1 
N  n 1
a
nN
n
is less than
aN
 n
.
NOTE: This theorem is stating that the error in approximating a convergent
alternating series is less than the absolute value of the first term omitted in the
summation.
Proof Will be provided later.

Example Approximate the sum of the convergent alternating series

(  1) n
 1
n 1
1
n4
to three decimal places.

You can verify that the alternating series
 (  1)
n  1
n 1
1
converges by the
n4
Alternating Series Test.
Recall, that in order for an approximate to be accurate to at least p decimal places,
 p
the error in the approximation must be less than 0.5  10 .
In order for our approximation to be accurate to at least three decimal places, we
3
need the error in the approximation to be less than 0.5  10 . Thus, we want
3
Error  0.5  10 .
n
Let a n  (  1 )
 1
1
.
n4
The theorem above states that the error in approximating a convergent alternating
series is less than the absolute value of the first term omitted in the summation.
That is, Error  a N  1 . Thus, we want to find the value of N such that
aN
 1
 0.5  10  3 . There are two ways to determine this value of N.
Method 1 Solve the inequality
aN
 1
 0.5  10  3 .
Copyrighted by James D. Anderson, The University of Toledo
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n  1
Since a n  (  1 )
1
, then
n4
aN
 1

1
. Thus,
( N  1) 4
aN
 1
 0.5  10  3 
1
 0.5  10  3  ( N  1 ) 4  2  10 3  ( N  1 ) 4  2000 
4
( N  1)
N  1
4
2000  N 
2000  1
2000  1  5.7 . Thus, N  6 . Thus, we need to

n  1 1
sum the first six terms in the series  (  1)
in order to obtain an
n4
n 1
approximate that has at least three decimals of accuracy. Thus,
Since
4

 (  1)
n 1
2000  6.7 , then
4
n  1
4
1
1
1
1
1
1
 1  4  4  4  4  4  0.947
4
n
2
3
4
5
6
Method 2 Find the first a N
calculation.
 1
3
which is less than 0.5  10  0.0005 by
a3 
1
 0.0123
34
a6 
1
 0.00077
64
a7 
1
 0.00042  0.0005
74
6
Thus, the first term that we omit in the summation is a 7  (  1 )
we sum the first six terms as we did above.
Answer: 0.947
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1
1

. Thus,
74
74
Example Approximate the sum of the convergent alternating series

1
(  1) n  1

to five decimal places.
( 3 n )!
n0

You can verify that the alternating series

(  1) n
n0
 1
1
converges by the
( 3 n )!
Alternating Series Test.
n
Let a n  (  1 )
 1
1
.
( 3 n )!
In order for our approximation to be accurate to at least five decimal places, we
5
need the error in the approximation to be less than 0.5  10 . Thus, we want
5
Error  0.5  10 .
The theorem above states that the error in approximating a convergent alternating
series is less than the absolute value of the first term omitted in the summation.
That is, Error  a N  1 . Thus, we want to find the value of N such that
aN
 1
 0.5  10  5 . In order to determine this value of N we will find the first
aN
 1
5
which is less than 0.5  10  0.000005 by calculation.
a2 
1
 0.0013
6!
a3 
1
 0.0000028  0.000005
9!
4
Thus, the first term that we omit in the summation is a 3  (  1)
1
1
 . Thus, we
9! 9!
will sum the first three terms since the summation starts at n  0 . Thus,


n0
(  1) n
 1
1
1
1
 1 

  0.83472
( 3 n )!
3!
6!
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer:  0.83472
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860