INFINITE SEQUENCES AND SERIES
The convergence tests that we have
looked at so far apply only to series
with positive terms.
12.5Alternating Series
In this section, we will learn:
How to deal with series
whose terms alternate in sign.
INFINITE SEQUENCES AND SERIES
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 1 1 ( 1)1 ...2 3 4 5 6
1 2 3 4 5 6... ( 1)
2 3 4 5 6 7 1
n
n
n
n
n
n
n
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
satisfies
i. bn+1 ≤ bn for all n
ii.
then the series is convergent.
11 2 3 4 5 6
1
( 1) ...
0
nn
n
n
b b b b b b b
b
lim 0nnb
ALTERNATING SERIES
Before giving the proof, let’s look at this figure
—which gives a picture of the idea behind the
proof.
Fig. 12.5.1, p. 746
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.
Fig. 12.5.1, p. 746
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.
Fig. 12.5.1, p. 746
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.
Fig. 12.5.1, p. 746
ALTERNATING SERIES
The even partial sums s2, s4, s6, . . .
are increasing.
The odd partial sums s1, s3, s5, . . .
are decreasing.
Fig. 12.5.1, p. 746
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,
Now, we compute the limit of the odd partial sums:
2lim nns s
2 1 2 2 1
2 2 1
lim lim( )
lim lim
0 (condition ii)
n n nn n
n nn n
s s b
s b
s
s
ALTERNATING SERIES TEST—PROOF
As both the even and odd partial sums
converge to s, we have
See Exercise 80(a) in Section 11.1
Thus, the series is convergent.
lim nns s
ALTERNATING SERIES
The alternating harmonic series
satisfies
i. bn+1 < bn because
ii.
It is convergent by the Alternating Series Test.
Example 1
1
1
1 1 1 ( 1)1 ...2 3 4
n
n n
1 1
1n n
1lim lim 0nn nb
n
ALTERNATING SERIES
The figure illustrates Example 1 by showing
the graphs of the terms an = (–1)n – 1/n
and the partial sums sn.
Fig. 12.5.2, p. 747
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).
Fig. 12.5.2, p. 747
ALTERNATING SERIES
The series is alternating.
However,
So, condition ii is not satisfied.
Example 2
1
( 1) 3
4 1
n
n
n
n
3 3 3lim lim lim
14 1 44n
n n n
nb
nn
ALTERNATING SERIES
Instead, we look at the limit of the nth term
of the series:
This limit does not exist.
So, the series diverges by the Test for Divergence.
Example 2
( 1) 3lim lim
4 1
n
nn n
na
n
ALTERNATING SERIES
Test the series
for convergence or divergence.
The given series is alternating.
So, we try to verify conditions i and ii of the Alternating Series Test.
Example 3
21
31
( 1)1
n
n
n
n
ALTERNATING SERIES
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:
Example 3
3
3 2
(2 )'( )
( 1)
x xf x
x
ALTERNATING SERIES
Since we are considering only positive x,
we see that f’(x) < 0 if 2 – x3 < 0, that is,
x > .
Thus, f is decreasing on the interval ( , ∞).
Example 3
3 2
3 2
ALTERNATING SERIES
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:
Thus, the given series is convergent by the Alternating Series Test.
2
3
3
1
lim lim lim 011 1
nn n n
n nbn
n
Example 3
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.
then |Rn| = |s – sn| ≤ bn+1
lim 0nnb
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
correct to three decimal places.
By definition, 0! = 1.
Example 4
0
( 1)
!
n
n n
ESTIMATING SUMS
First, we observe that the series is convergent
by the Alternating Series Test because:
i.
ii.
1 1 1
( 1)! !( 1) !n n n n
1 1 10 0 so 0 as
! !n
n n 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:
1 1 1 1 1 12 6 24 120 720 5040
1 1 1 1 1 1 1 1...
0! 1! 2! 3! 4! 5! 6! 7!1 1 ...
s
Example 4
ESTIMATING SUMS
Notice that
and
1 17 5040 5000 0.0002b
1 1 1 1 16 2 6 24 120 7201 1
0.368056
s
1 1 1 1 1 12 6 24 120 720 5040
1 1 1 1 1 1 1 1...
0! 1! 2! 3! 4! 5! 6! 7!1 1 ...
s
Example 4
ESTIMATING SUMS
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.
Example 4