solutions for quiz xxix

In short ln r 1 hence we conclude r e 9 1 2n2 1 3n3

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Unformatted text preview: · 2 3 4 n 2n 3n 4n 5n5 1 1 1 1 1 − + − + − ··· 2 3 4 n 2n 3n 4n 5n5 8 = lim n− ∞ → −n− 1 2 − 1 3n − 1 4n2 − 1 5n3 − ··· n− 1 2 + 1 3n − 1 4n2 + 1 5n3 − ··· − 1 n + 2 3n2 − 1 2n3 + ··· + + 2 1 n + = lim n− ∞ → 1− 1 3n2 1 6n3 − = 1. In short, ln R = 1. Hence we conclude R = e. 9 − − ··· 1 2n2 + 1 3n3 − ··· ∞ (2) x2 The radius of convergence for n is R = 1. n=0 : Work Assume x > 0. Consider ∞ x2 (♭) n x + x2 = + x4 + x8 +··· n=0 ∞ xn = (♭♭) n=0 1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + · · · First, if x = 1, then (♭) is the sum of infinitely many replicas of 1, hence it clearly diverges. Next, if (♭♭) converges then (♭) converges by comparison test . In particular, if x < 1 then then (♭) converges . By the definition of the radius of convergence, if there exis...
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This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

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