11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
The convergence tests that we have
looked at so far apply only to series
with positive terms.
INFINITE SEQUENCES AND SERIES
11.5
Alternating Series
In this section, we will learn:
How to deal with series
whose terms alternate in sign.
ALTERNATING SERIES
An alternating series is a series whose
terms are alternately positive and negative.
 Here are two examples:
1
1

2

1
2

1

1
3
2
3


4
3
4

1

5
4
5

1

 ... 
6
5
6


(  1)
n
n 1
6
7
n 1

 ... 
 (  1)
n 1
n
n
n 1
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|
ALTERNATING SERIES
The following test states that, if the terms
of an alternating series decrease toward 0
in absolute value, the series converges.
ALTERNATING SERIES TEST
If the alternating series

 (  1)
n 1
b n  b1  b 2  b3  b 4  b5  b6  ...
n 1
satisfies
i. bn+1 ≤ bn for all n
ii. lim b n  0
n 
then the series is convergent.
bn  0
ALTERNATING SERIES
Before giving the proof, let’s look at this
figure—which gives a picture of the idea
behind the proof.
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.
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.
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.
ALTERNATING SERIES
The even partial sums s2, s4, s6, . . .
are increasing.
The odd partial sums s1, s3, s5, . . .
are decreasing.
ALTERNATING SERIES
Thus, it seems plausible that both
are converging to some number s,
which is the sum of the series.
 So, we consider the even and odd partial sums
separately in the following proof.
ALTERNATING SERIES TEST—PROOF
First, we consider the even partial sums:
s2 = b1 – b2 ≥ 0
since b2 ≤ b1
s4 = s2 + (b3 – b4) ≥ s2
since b4 ≤ b3
ALTERNATING SERIES TEST—PROOF
In general,
s2n = s2n – 2 + (b2n – 1 – b2n) ≥ s2n – 2
since b2n ≤ b2n – 1
Thus,
0 ≤ s2 ≤ s4 ≤ s6 ≤ … ≤ s2n ≤ …
ALTERNATING SERIES TEST—PROOF
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.
ALTERNATING SERIES TEST—PROOF
Thus, the sequence {s2n} of even
partial sums is increasing and bounded
above.
 Therefore, it is convergent by the Monotonic
Sequence Theorem.
ALTERNATING SERIES TEST—PROOF
Let’s call its limit s, that is,
lim s 2 n  s
n 
 Now, we compute the limit of the odd partial sums:
lim s 2 n  1  lim ( s 2 n  b 2 n  1 )
n 
n 
 lim s 2 n  lim b 2 n  1
n 
 s0
 s
n 
(condition ii)
ALTERNATING SERIES TEST—PROOF
As both the even and odd partial sums
converge to s, we have lim s n  s
n 
 See Exercise 80(a) in Section 11.1
Thus, the series is convergent.
Example 1
ALTERNATING SERIES
The alternating harmonic series
1
1

2
1

3
1
4

 ... 

n 1
(  1)
n 1
n
satisfies
i. bn+1 < bn
ii. lim bn  lim
n 

n 
because
1
0
1
n 1

1
n
n
It is convergent by the Alternating Series Test.
ALTERNATING SERIES
The figure illustrates Example 1 by showing
the graphs of the terms an = (–1)n – 1/n
and the partial sums sn.
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
(Exercise 36).
Example 2
ALTERNATING SERIES
The series


n 1
(  1) 3 n
n
is alternating.
4n  1
However,
lim b n  lim
n 
n 
3n
4n  1
 lim
n 
3
4
1
n
 So, condition ii is not satisfied.

3
4
Example 2
ALTERNATING SERIES
Instead, we look at the limit of the nth term
of the series:
(  1) 3 n
n
lim a n  lim
n 
n 
4n  1
This limit does not exist.
 So, the series diverges by the Test for Divergence.
Example 3
ALTERNATING SERIES
Test the series

 (  1)
n 1
n 1

n
2
n 1
3
for convergence or divergence.
 The given series is alternating.
 So, we try to verify conditions i and ii
of the Alternating Series Test.
ALTERNATING SERIES
Example 3
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:
x (2  x )
3
f '( x ) 
( x  1)
3
2
ALTERNATING SERIES
Example 3
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 , ∞).
ALTERNATING SERIES
Example 3
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.
Example 3
ALTERNATING SERIES
Condition ii is readily verified:
1
lim b n  lim
n 
n 
n
2
n 1
3
 lim
n 
n
1
1
n
 Thus, the given series is convergent
by the Alternating Series Test.
3
0
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.
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.
ALTERNATING SERIES ESTIMATION THEOREM
If s = Σ (–1)n-1bn is the sum of an alternating
series that satisfies
i. 0 ≤ bn+1 ≤ bn
ii. lim b n  0
n 
then
|Rn| = |s – sn| ≤ bn+1
ALTERNATING SERIES ESTIMATION THM.—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
ALTERNATING SERIES ESTIMATION THEOREM
You can see geometrically why the theorem
is true by looking at this figure.
 Notice that s – s4 < b5, |s – s5| < b6, and so on.
 Notice also that s lies between any two consecutive
partial sums.
ESTIMATING SUMS
Find the sum of the series
Example 4


(  1)
n0
correct to three decimal places.
 By definition, 0! = 1.
n!
n
Example 4
ESTIMATING SUMS
First, we observe that the series is convergent
by the Alternating Series Test because:
i.
1
( n  1) !
ii. 0 
1
n!


1
n
1
n !( n  1)
 0
1

so
n!
1
n!
 0 as n  
Example 4
ESTIMATING SUMS
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:
s
1
0!

1

1!
 11
1
2
1

2!

1
6
1

3!

1
24

1

4!
1
120
1

5!

1
720
1

6!

1
5040
1
7!
 ...
 ...
Example 4
ESTIMATING SUMS
s
1
0!

1

1!
 11
1

2!
1
2

1
6
1
1

3!

 Notice that b7 
and s 6  1  1 
1
120

 0.368056
1
6

5!


1
5040
1
2

4!

1
24
1

1
720
1
5000
1
24
1

6!

1
7!
1
5040
 ...
 0.0002

1
120

1
720
 ...
ESTIMATING SUMS
Example 4
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.
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.