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Unformatted text preview: , but the interval of convergence is translated to the right by c when
as before by R = lim
n → ∞ an+1
0 < R < ∞, and now is one of
(−R + c, R + c) , [−R + c, R + c) , (−R + c, R + c] , Example 3: ﬁnd the radius of 3
c= ,
2 convergence of the power series
∞
n=1 n2
(2x − 3)n .
4n an n2
= n.
2 But then
n
n2 2n+1
an
=2
=
2 2n
an+1
(n + 1)
n+1 2 . Consequently, Solution: since
(2x − 3)n = 2n x −
this is a power series [−R + c, R + c] . ∞
n=0 3
,
2 an (x − c)n with R = lim n→∞ an
n
= lim 2
n→∞
an+1
n+1 2 = 2. Example 4: ﬁnd the interval of
3
3
− 2, + 2 ,
2
2 convergence of the power series
∞
n=1 n2
(2x − 3)n .
4n 17
.
−,
22 i.e., outside We have to determine whether the series
converges at x = −...
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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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