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

714 Chapter 11 Infinite Sequences and Series
The Test for Divergence follows from Theorem 6 because, if the series is not divergent,
then it is convergent, and so lim n l ` a n − 0.
Example 10 Show that the series ò
n−1
SOLUTION
n 2
lim
nl` an − lim
nl` 5n 2 1 4 − lim
nl`
n 2
5n 2 1 4 diverges.
1
5 1 4yn 2 − 1 5 ± 0
So the series diverges by the Test for Divergence.
n
Note 3 If we find that lim n l ` a n ± 0, we know that o a n is divergent. If we find
that lim n l ` a n − 0, we know nothing about the convergence or divergence of o a n .
Remember the warning in Note 2: if lim n l ` a n − 0, the series o a n might converge or it
might diverge.
8 Theorem If o a n and o b n are convergent series, then so are the series o ca n
(where c is a constant), osa n 1 b n d, and osa n 2 b n d, and
(i) ò ca n − c ò a n
n−1
n−1
(ii) ò sa n 1 b n d − ò a n 1 ò b n
n−1
n−1 n−1
(iii) ò sa n 2 b n d − ò a n 2 ò b n
n−1
n−1 n−1
These properties of convergent series follow from the corresponding Limit Laws for
Sequences in Section 11.1. For instance, here is how part (ii) of Theorem 8 is proved:
Let
s n − o n
t n − o n
a i
i−1
s − ò a n
n−1
The nth partial sum for the series osa n 1 b n d is
and, using Equation 5.2.10, we have
u n − o n
i−1
sa i 1 b i d
b i
i−1
t − ò b n
n−1
lim
n l ` un − lim
n l ` on sa i 1 b i d − lim
i−1
n l `So n
a i 1 o n
b
i−1 i−1
iD
− lim
n l ` on a i 1 lim
i−1 n l ` on b i
i−1
− lim
n l ` sn 1 lim
n l ` tn − s 1 t
Therefore o sa n 1 b n d is convergent and its sum is
ò sa n 1 b n d − s 1 t − ò a n 1 ò b n
n−1
n−1 n−1
n
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