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

Section 11.2 Series 713
Example 9 Show that the harmonic series
is divergent.
1
ò
n−1 n − 1 1 1 2 1 1 3 1 1 4 1 ∙ ∙ ∙
SOLUtion For this particular series it’s convenient to consider the partial sums s 2 , s 4 ,
s 8 , s 16 , s 32 , . . . and show that they become large.
s 2 − 1 1 1 2
s 4 − 1 1 1 2 1 s 1 3 1 1 4 d . 1 1 1 2 1 s 1 4 1 1 4 d − 1 1 2 2
s 8 − 1 1 1 2 1 s 1 3 1 1 4 d 1 s 1 5 1 1 6 1 1 7 1 1 8 d
. 1 1 1 2 1 s 1 4 1 1 4 d 1 s 1 8 1 1 8 1 1 8 1 1 8 d
The method used in Example 9 for
showing that the harmonic series
diverges is due to the French scholar
Nicole Oresme (1323–1382).
− 1 1 1 2 1 1 2 1 1 2 − 1 1 3 2
s 16 − 1 1 1 2 1 s 1 3 1 1 4 d 1 s 1 5 1 ∙ ∙ ∙ 1 1 8 d 1 s 1 9 1 ∙ ∙ ∙ 1 1
16 d
. 1 1 1 2 1 s 1 4 1 1 4 d 1 s 1 8 1 ∙ ∙ ∙ 1 1 8 d 1 s 1
16 1 ∙ ∙ ∙ 1 1
16 d
− 1 1 1 2 1 1 2 1 1 2 1 1 2 − 1 1 4 2
Similarly, s 32 . 1 1 5 2 , s 64 . 1 1 6 2 , and in general
s 2 n . 1 1 n 2
This shows that s 2 n l ` as n l ` and so hs n j is divergent. Therefore the harmonic
series diverges.
n
6 Theorem If the series ò a n is convergent, then lim
n−1
n l ` an − 0.
Proof Let s n − a 1 1 a 2 1 ∙ ∙ ∙ 1 a n . Then a n − s n 2 s n21 . Since o a n is convergent,
the sequence hs n j is convergent. Let lim n l ` s n − s. Since n 2 1 l ` as n l `,
we also have lim n l ` s n21 − s. Therefore
lim
n l ` an − lim
n l `ss n 2 s n21 d − lim s n 2 lim s n21 − s 2 s − 0
n l ` n l `
n
Note 1 With any series o a n we associate two sequences: the sequence hs n j of
its partial sums and the sequence ha n j of its terms. If o a n is convergent, then the limit of
the sequence hs n j is s (the sum of the series) and, as Theorem 6 asserts, the limit of the
sequence ha n j is 0.
note 2 The converse of Theorem 6 is not true in general. If lim n l ` a n − 0, we cannot
conclude that o a n is convergent. Observe that for the harmonic series o1yn we have
a n − 1yn l 0 as n l `, but we showed in Example 9 that o1yn is divergent.
7 Test for Divergence If lim a n does not exist or if lim a n ± 0, then the
nl` nl`
series ò a n is divergent.
n−1
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