James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

734 Chapter 11 Infinite Sequences and Series
Figure 2 illustrates Example 1 by
show ing the graphs of the terms
a n − s21d n21 yn and the partial sums
s n. Notice how the values of s n 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 (see Exercise 36).
1
s n
Example 1 The alternating harmonic series
satisfies
1 2 1 2 1 1 3 2 1 4 1 ∙ ∙ ∙ − s21d
ò
n21
n−1 n
(i) b n11 , b n
because
1
(ii) lim b n − lim
nl` nl` n − 0
1
n 1 1 , 1 n
so the series is convergent by the Alternating Series Test.
n
a n
0 n
s21d
Example 2 The series ò
n 3n
is alternating, but
n−1 4n 2 1
3n
lim
n l ` bn − lim
n l ` 4n 2 1 − lim
nl`
3
4 2 1 n
− 3 4
FIGURE 2
so condition (ii) is not satisfied. Instead, we look at the limit of the nth term of the series:
s21d n 3n
lim
n l ` an − lim
n l ` 4n 2 1
This limit does not exist, so the series diverges by the Test for Divergence.
n
Example 3 Test the series ò s21d n11
n−1
n 2
for convergence or divergence.
n 3 1 1
SOLUTION The given series is alternating so we try to verify conditions (i) and (ii) of
the Alternating Series Test.
Unlike the situation in Example 1, it is not obvious that the sequence given by
b n − n 2 ysn 3 1 1d is decreasing. However, if we consider the related function
f sxd − x 2 ysx 3 1 1d, we find that
f 9sxd − xs2 2 x 3 d
sx 3 1 1d 2
Instead of verifying condition (i) of
the Alternating Series Test by computing
a derivative, we could verify that
b n11 , b n directly by using the tech nique
of Solution 1 of Example 11.1.13.
Since we are considering only positive x, we see that f 9sxd , 0 if 2 2 x 3 , 0, that is,
x . s 3 2 . Thus f is decreasing on the interval ss 3 2 , `d. This means that f sn 1 1d , f snd
and therefore b n11 , b n when n > 2. (The inequality b 2 , b 1 can be verified directly
but all that really matters is that the sequence hb n j is eventually decreasing.)
Condition (ii) is readily verified:
n 2
lim
n l ` bn − lim
n l ` n 3 1 1 − lim
n l `
1
n
1 1 1 n 3 − 0
Thus the given series is convergent by the Alternating Series Test.
n
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