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

Section 11.4 The Comparison Tests 727
43. (a) Use (4) to show that if s n is the nth partial sum of the
harmonic series, then
s n < 1 1 ln n
(b) The harmonic series diverges, but very slowly. Use
part (a) to show that the sum of the first million terms is
less than 15 and the sum of the first billion terms is less
than 22.
44. Use the following steps to show that the sequence
t n − 1 1 1 2 1 1 3 1 ∙ ∙ ∙ 1 1 n 2 ln n
has a limit. (The value of the limit is denoted by and is
called Euler’s constant.)
(a) Draw a picture like Figure 6 with f sxd − 1yx and interpret
t n as an area [or use (5)] to show that t n . 0 for all n.
(b) Interpret
t n 2 t n11 − flnsn 1 1d 2 ln ng 2 1
n 1 1
as a difference of areas to show that t n 2 t n11 . 0. Therefore
ht nj is a decreasing sequence.
(c) Use the Monotonic Sequence Theorem to show that ht nj
is convergent.
ln n
45. Find all positive values of b for which the series o ǹ−1 b
converges.
46. Find all values of c for which the following series converges.
n−1S ò
c n 2 1
n 1 1D
In the comparison tests the idea is to compare a given series with a series that is known
to be convergent or divergent. For instance, the series
1
ò
n−1
1
2 n 1 1
reminds us of the series oǹ−1 1y2 n , which is a geometric series with a − 1 2 and r − 1 2 and
is therefore convergent. Because the series (1) is so similar to a convergent series, we
have the feeling that it too must be convergent. Indeed, it is. The inequality
1
2 n 1 1 , 1 2 n
shows that our given series (1) has smaller terms than those of the geometric series and
therefore all its partial sums are also smaller than 1 (the sum of the geometric series).
This means that its partial sums form a bounded increasing sequence, which is convergent.
It also follows that the sum of the series is less than the sum of the geometric series:
ò
n−1
1
2 n 1 1 , 1
Similar reasoning can be used to prove the following test, which applies only to series
whose terms are positive. The first part says that if we have a series whose terms are
smaller than those of a known convergent series, then our series is also convergent. The
second part says that if we start with a series whose terms are larger than those of a
known divergent series, then it too is divergent.
The Comparison Test Suppose that o a n and o b n are series with positive terms.
(i) If o b n is convergent and a n < b n for all n, then o a n is also convergent.
(ii) If o b n is divergent and a n > b n for all n, then o a n is also divergent.
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