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Unformatted text preview: Example
Consider the following two inﬁnite series:
∞ (a)
n=0 ∞ 2 3n−1 (b) 9n n=1 (−7)n . For each series, ﬁnd its sum and tell whether it converges or diverges. Solution:
Each of the series in (a) and (b) is a geometric series. Recall that a geometric series may be
written as
∞
∞
arn arn−1 or n=0 n=1 (the two summations are equivalent), and that if r < 1, the series converges with sum given
by
∞
∞
a
n
.
ar =
arn−1 =
1−r
n=0
n=1
Otherwise, if r ≥ 1, the geometric series diverges.
∞ (a)
n=0 2
9n Because the summation for this series begins with the n = 0 term, it is natural to try to
∞ arn . Since express it in the form
n=0 n=0 2
=
9n ∞ 2·
n=0 1
9 n , it must be that 1
r= ,
9
so that r < 1. Thus the series converges and its sum is given by
a = 2, ∞ 2·
n=0 1
9 n = 2
1− 1
9 = 2
8
9 = 18
9
=.
8
4 ∞ (b)
n=1 3 n −1
(−7)n . The summation for this series begins with the n = 1 term, so it is natural to try to express
∞ arn−1 . This means that all powers inside the summation need to the series in the form
n=1 be of the form (n − 1). We have
∞ n=1 3n−1
=
(−7)n ∞ n=1
∞ =
n=1
∞ 3n−1
(−7)1 · (−7)n−1
1
3n−1
·
(−7)1 (−7)n−1
− =
n=1 1
7 − 3
7 n−1 ∞ arn−1 with so that the series is now in the form of
n=1 1
a=− ,
7 3
r=− .
7 Again, r < 1 so the geometric series converges, and its sum is given by
∞ n=1 1
−
7 3
−
7 n−1 = −1/7
−1/7
1
=
=− .
1 − (−3/7)
10/7
10 2 ...
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus, Geometric Series

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