Vanier College
Calculus 2 (201-NYB-05)
Section: 007, Semester: Winter 2011
§11.5 Alternating Series (p710)
1. Alternating Series
Definition of Alternating Series (p710)
An alternating series is a series whose terms are alternately positive and negative.

n 1
  1 bn
n 1
(bn  0) or

  1 b
n
n
n 1
(bn  0)
The Alternating Series Test (p710)

If the alternating series
  1
n 1
n 1
(1)
(2)
bn+1  bn
lim bn  0
(bn  0) satisfies
bn
for all n
n 
then the series is convergent.
Proof Consider the even partial sums:
s2  b1  b2  0
s 4  s 2  b3  b4  s 2
In general,
s 2m  s 2m2  (b2m1  b2 m )  s 2 m2
since b2  b1
since b4  b3
since b2m  b2m-1
Thus 0  s 2  s 4    s 2 m  
But we can write
s 2 m  b1  (b2  b3 )  (b4  b3 )    (b2 m2  b2 m1 )  b2m  b1 for all n, since every
term in bracket is positive.
By the Monotonic sequence Theorem the sequence {s2m} is convergent.
Let lim s 2 m  s
(1)
m 
lim s 2 m 1  lim ( s 2 m  b2 m 1 )  lim s 2 m  lim b2 m 1  s  0  s
m 
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m 
n 
n 
1
(2)
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Let  >0.
From (1), there exists N1, such that a2m  L   for m > N1
(Definition of the limit of the sequence 11.1.2, p677)
From (2), there exists N2, such that a2m1  L   for m > N2
(Definition of the limit of the sequence 11.1.2, p677)
Let N=max{2N1, N2 + 1} and n > N.
If n is even, that is n = 2m where m > N1, so an  L  a2m  L  
If n is odd, that is n = 2m+1 where m > N2, so an  L  a2m1  L  
Therefore by the Definition of the limit of the sequence 11.1.2, lim s n  s
n
Example 1
(p711)
(1) n 1
for convergence or divergence:
n
n 1

Test the alternating harmonic series 
Sol.
Let bn 
1
.
n
We have bn 1 
1
1
 bn 
n 1
n
1
0
n 
n  n
So the series is convergent by the Alternating Series Test.
lim bn  lim
Example 2
(p711)
(1) 3n
for convergence or divergence:
n 1 4n  1

Test the

n
Sol.
3n
.
4n  1
3n
3
lim bn  lim
  0
n 
n  4n  1
4
This limit does not exist, so the series is divergent by the Test for Divergence..
Let bn 
Example 3
(p712)
Test the
n 1
(1) n
for convergence or divergence:
n3  1
n 1


2
Sol.
x2
Let f ( x)  3
.
x 1
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2
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f ( x) 
x(2  x 3 )
( x 3  1) 2
f ( x)  0 if 2 – x3 < 0, that is, x  3 2
Thus f(x) is decreasing on (3 2 , )
Then f (n  1)  f (n) , and so bn+1 < bn when n  2.
n2
 0
n  n 3  1
By the Alternating Series Test, the series is convergent.
lim
2. Estimating Sums
(p712)
Alternating Series Estimation Theorem (p712)

If the alternating series
  1
n 1
(1)
(2)
bn+1  bn
lim bn  0
n 1
(bn  0) satisfies
bn
for all n
n 
then | Rn | = | s – sn |  bn+1.
Proof
Since the alternating series satisfies the hypotheses of the alternating series test,

then the series
  1
n 1
n 1
bn is convergent.
Suppose lim s n  s .
n
R2 m  s  s 2 m
 (b2 m 1  b2 m  2 )  (b2 m 3  b2 m 5 )  
 b2 m 1  b2 m  2   (b2 m 3  b2 m  4 )  
So
R2m  b2m1  (b2m2  b2m3 )  b2m4  b2m5     b2m1
Similarly,
R2m1  (b2 m  b2 m1  b2 m 2  b2 m3  )  0 ,
And so,
R2m1  b2m  (b2m1  b2m2 )  b2m3  b2m4     b2m
From (1) and (2), we have
Rn  bn1
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3
(1)
(2)
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Example 4
(p713)
 1n correct to three decimal places (By definition,
Find the sum of the series 
n!
n 0
0! = 1)

Sol.
Let bn 
1
.
n!
We have bn 1 
1
1
1

  bn
(n  1)! n  1n! n!
1 1

n! n
1
And  0 as n  
n
By the Squeeze Theorem, the series is convergent.
1
lim bn  lim  0
n 
n   n!
So the series is convergent by the Alternating Series Test.
0
1
n  1!
1
1
1
1
1
 0.001389 , b7 

 0.0002 .
Notice b5  
, b6 
5! 120
720
5040 5000
This error of less than 0.0002 does not affect the third decimal place. Then choose
1 1 1
1
 0.368
n = 6, we have s 6      
0! 1! 2!
720
By the Alternating Series Estimation Theorem, Rn  bn 1 
Exercises 11.5:
5,
7,
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9,
(p694)
11,
13,
15,
4
17,
19,
25,
29,
32
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