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Unformatted text preview: n+3 n=0 (−1)n √ 1
,
n+3 which converges by the Alternating series test,
while at x = − 1 the series becomes
4 Solution: this is the same power series
∞
n=0 (−4)n n
√
x=
n+3 ∞
n n=0 (−1) √ 4n
xn
n+3 we looked at in Example 1. There we saw that
R = 1 , so the series converges for x in the
4
1
interval (− 4 , 1 ) but diverges for x outside
4 [− 1 , 1 ]. We have to determine whether the
44
1
series converges at x = ± 4 . ∞
n=0 √ 1
,
n+3 which diverges by the Limit Comparison test
1
because
diverges. Thus the interval
1/2
nn
of convergence of the series is
11
−,
44 . Sometimes it’s very useful to center things away from the origin, say at x = c, and work with power
series
∞
2 n 3 a0 + a1 (x − c) + a2 (x − c) + a3 (x − c) + . . . + an (x − c) + . . . = n=0 an (x − c)n . We’ll say such a series is a Power series centered at x = c. Its Radius of convergence is deﬁned exactly
an...
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This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Gilbert
 Power Series

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