solutions for quiz xxix

1 1 32 n 1 32n2 1 42 n 1 42n2 10 1 52 n 1 52n2 1 62

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Unformatted text preview: ts a positive real number R such that a power series with variable x diverges for x = R and converges for x with 0 < x < R, then the radius of convergence for the same power series is R. In our case, R = 1. [III] (Extra 10pts) ζ (1) lim n− ∞ → = lim n− ∞ → Let ζ s 2n −1 ζ 2n + 2 1 22 n 1 22n+2 denote the Riemann’s zeta function. −1 + + 1 32 n 1 32n+2 + + 1 42 n 1 42n+2 10 + + 1 52 n 1 52n+2 + + 1 62 n 1 62n+2 + ··· +··· = 2 3 1 1 + 22 n lim n− ∞ → 1 22n+2 1 + lim 4 n− ∞ → · 1 + 1+ = 4 · 1+ = 4 lim n− ∞ → lim n− ∞ → 1+0 1+0 · = 2n 2 3 + 2n+2 2 3 2n 2 3 + 2n+2 2 4 2n+2 + 2n 2 4 + 2 5 + 2n+2 + + 2 5 + 2n 2n+2 2 5 2n 2 4 2 4 + 2 5 + 2n+2 2 4 + 2 3 1 + 2n 2 4 + 2n+2 2 3 = 2n 2n + 2n+2 + 2 5 2 5 + + 2n 2n+2 + + 2 6 2n 2 6 2n+2 2 6 2 6 + ··· + ··· 2n + ··· 2n+2 + ··· 2 6 2n 2 6 2n+2 + ··· lim n− ∞ → lim n− ∞ → 2 3 2n 4. 2 3 2n+2 + + 2 4 2n 2 4 2n+2 2 5 2n 2 5 2n+2 + + + + These are true because of the squeeze theorem. Indeed, 11 2 6 2 6 2n + · · · = 0, 2n+2 + ··· In the above, the following are used: + · · · = 0. 0...
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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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