There we saw that r 1 so the series converges for x

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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.

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