4n an n2 n 2 but then n n2 2n1 an 2 2 2n an1 n

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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.

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