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

Section 11.6 Absolute Convergence and the Ratio and Root Tests 739
The Ratio Test
(i) If lim
nl`
Z
a n11
a n
(and therefore convergent).
Z − L , 1, then the series ò a n is absolutely convergent
n−1
(ii) If lim
nl`
Z
a n11
a n
divergent.
Z − L . 1 or lim
nl`
Z
a n11
a n
Z − `, then the series ò a n is
n−1
(iii) If lim
nl`
Z
a n11
a n
Z − 1, the Ratio Test is inconclusive; that is, no conclusion
can be drawn about the convergence or divergence of o a n .
Proof
(i) The idea is to compare the given series with a convergent geometric series. Since
L , 1, we can choose a number r such that L , r , 1. Since
a
lim Z
n11
nl` a n
Z − L and L , r
the ratio | a n11ya n | will eventually be less than r; that is, there exists an integer N
such that
or, equivalently,
Z
a n11
a n
Z
, r whenever n > N
4
| a n11 | , | a n | r whenever n > N
Putting n successively equal to N, N 1 1, N 1 2, . . . in (4), we obtain
and, in general,
5
| a N11 | , | a N | r
| a N12 | , | a N11 | r , | a N | r 2
| a N13 | , | a N12 | r , | a N | r 3
| a N1k | , | a N | r k for all k > 1
Now the series
ò
k−1 | a N | r k − | a N | r 1 | a N | r 2 1 | a N | r 3 1 ∙ ∙ ∙
is convergent because it is a geometric series with 0 , r , 1. So the inequality (5),
together with the Comparison Test, shows that the series
ò | a n | − ò | a N1k | − | a N11 | 1 | a N12 | 1 | a N13 | 1 ∙ ∙ ∙
n−N11 k−1
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