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

Section 11.3 The Integral Test and Estimates of Sums 721
equal to the integral y`
1 (1ysx ) dx. But we know from Section 7.8 that this improper
integral is divergent. In other words, the area under the curve is infinite. So the sum of
the series must be infinite; that is, the series is divergent.
The same sort of geometric reasoning that we used for these two series can be used to
prove the following test. (The proof is given at the end of this section.)
The Integral Test Suppose f is a continuous, positive, decreasing function on
f1, `d and let a n − f snd. Then the series oǹ−1 an is convergent if and only if the
improper integral y`
1
f sxd dx is convergent. In other words:
(i) If y`
1
(ii) If y`
1
f sxd dx is convergent, then ò a n is convergent.
n−1
f sxd dx is divergent, then ò a n is divergent.
n−1
Note When we use the Integral Test, it is not necessary to start the series or the
integral at n − 1. For instance, in testing the series
ò
n−4
1
sn 2 3d we use y`
2 4
1
sx 2 3d 2 dx
Also, it is not necessary that f be always decreasing. What is important is that f be
ultimately decreasing, that is, decreasing for x larger than some number N. Then o ǹ−N an
is convergent, so o ǹ−1 an is convergent by Note 4 of Section 11.2.
Example 1 Test the series ò
n−1
1
for convergence or divergence.
n 2 1 1
SOLUtion The function f sxd − 1ysx 2 1 1d is continuous, positive, and decreasing on
f1, `d so we use the Integral Test:
y`
1
1
dx − lim
x 2 1 1 t l ` yt 1
1
t
dx − lim xg
x 2 1 1 t l ` tan21 1
− lim
t l `Stan 21 t 2 4D − 2 2 4 − 4
Thus y`
1 1ysx 2 1 1d dx is a convergent integral and so, by the Integral Test, the series
o 1ysn 2 1 1d is convergent.
n
In order to use the Integral Test we
need to be able to evaluate y`
1 f sxd dx
and therefore we have to be able to find
an antiderivative of f . Frequently this
is difficult or impossible, so we need
other tests for convergence too.
Example 2 For what values of p is the series ò
n−1
1
n p convergent?
SOLUtion If p , 0, then lim n l ` s1yn p d − `. If p − 0, then lim n l ` s1yn p d − 1. In
either case lim n l ` s1yn p d ± 0, so the given series diverges by the Test for Divergence
(11.2.7).
If p . 0, then the function f sxd − 1yx p is clearly continuous, positive, and decreasing
on f1, `d. We found in Chapter 7 [see (7.8.2)] that
y`
1
1
p
dx converges if p . 1 and diverges if p < 1
x
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