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

712 Chapter 11 Infinite Sequences and Series
tec Module 11.2 explores a series
that depends on an angle in a triangle
and enables you to see how
rapidly the series converges when
varies.
Notice that the terms cancel in pairs.
This is an example of a telescoping
sum: Because of all the cancellations,
the sum collapses (like a pirate’s collapsing
telescope) into just two terms.
Thus
ò x n − 1 1 x 1 x 2 1 x 3 1 x 4 1 ∙ ∙ ∙
n−0
This is a geometric series with a − 1 and r − x. Since | r | − | x |
and (4) gives
5
Example 8 Show that the series ò
n−1
ò x n − 1
n−0 1 2 x
1
nsn 1 1d
, 1, it converges
is convergent, and find its sum.
SOLUtion This is not a geometric series, so we go back to the definition of a convergent
series and compute the partial sums.
s n − o n
i−1
1
isi 1 1d − 1
1 ? 2 1 1
2 ? 3 1 1
3 ? 4 1 ∙ ∙ ∙ 1 1
nsn 1 1d
We can simplify this expression if we use the partial fraction decomposition
(see Section 7.4). Thus we have
s n − o n
i−1
1
isi 1 1d i−1S − o n 1 i
1
isi 1 1d − 1 i
2 1
i 1 1D
2 1
i 1 1
−S1 2 1 2D 1S 1 2 2 1 3D 1S 1 3 2 1 4D 1 ∙ ∙ ∙ 1S 1 n 2 1
n 1 1D
− 1 2 1
n 1 1
n
and so
lim s n − lim 2 1 − 1 2 0 − 1
nl` nl`S1
n 1 1D
Therefore the given series is convergent and
ò
n−1
1
nsn 1 1d − 1
n
Figure 3 illustrates Example 8 by showing
the graphs of the sequence of terms
a n − 1y[nsn 1 1d] and the sequence
hs nj of partial sums. Notice that a n l 0
and s n l 1. See Exer cises 78 and 79
for two geometric interpretations of
Example 8.
1
s n
a n
FIGURE 3
0
n
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