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

722 Chapter 11 Infinite Sequences and Series
It follows from the Integral Test that the series o 1yn p converges if p . 1 and diverges
if 0 , p < 1. (For p − 1, this series is the harmonic series discussed in Example
11.2.9.) n
The series in Example 2 is called the p-series. It is important in the rest of this chapter,
so we summarize the results of Example 2 for future reference as follows.
1 The p-series ò
n−1
1
p
is convergent if p . 1 and divergent if p < 1.
n
Example 3
(a) The series
ò
n−1
1
n 3 − 1 1 3 1 1 2 3 1 1 3 3 1 1 4 3 1 ∙ ∙ ∙
is convergent because it is a p-series with p − 3 . 1.
(b) The series
ò
n−1
1
n 1y3
− ò
n−1
1
− 1 1 1 1 1 1 1 1 ∙ ∙ ∙
s 3 n s 3 2 s 3 3 s 3 4
is divergent because it is a p-series with p − 1 3 , 1.
n
Note We should not infer from the Integral Test that the sum of the series is equal to
the value of the integral. In fact,
ò
n−1
1
n − 2
2 6
whereas
y`
1
1
x 2 dx − 1
Therefore, in general,
ò a n ± y`
f sxd dx
n−1 1
ln n
Example 4 Determine whether the series ò converges or diverges.
n−1 n
SOLUtion The function f sxd − sln xdyx is positive and continuous for x . 1 because
the logarithm function is continuous. But it is not obvious whether or not f is decreasing,
so we compute its derivative:
f 9sxd −
s1yxdx 2 ln x
x 2
− 1 2 ln x
x 2
Thus f 9sxd , 0 when ln x . 1, that is, when x . e. It follows that f is decreasing
when x . e and so we can apply the Integral Test:
y`
1
ln x
x
dx − lim
t l ` y t ln x sln xd 2 t
dx − lim
1 x
t l ` 2
G1
sln td 2
− lim
t l ` 2
− `
Since this improper integral is divergent, the series o sln ndyn is also divergent by the
Integral Test.
n
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