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Unformatted text preview: Solutions to homework # 7. 1. (a) lim sup n →∞ n √ n 3 = lim n →∞ e 3 n log n = e = 1, so the radius of convergence is equal to 1 according to Theorem 3.39. (b) Theorem 3.39 can be applied here too, but this requires the fact lim n →∞ 1 n √ n ! = 0, which needs justification. Alternatively, one can apply the ratio test to prove that the radius of convergence is equal to ∞ . Indeed, for any fixed number z , lim n →∞ (2 z ) n +1 ( n + 1)! · n ! (2 z ) n = lim n →∞ 2 z n + 1 = 0 . (c) lim sup n →∞ n q 2 n /n 2 = 2 / lim n →∞ n 2 /n = 2, since lim n →∞ n 2 /n = exp(lim n →∞ 2 log n n ) = e = 1, so the radius of convergence is equal to 1 / 2 by Theorem 3.39. (d) lim sup n →∞ n q n 3 / 3 n = lim n →∞ n √ n 3 / 3 = 1 / 3 by the same reasoning as in (c), hence the radius of convergence is equal to 3 by Theorem 3.39. 2. (a) First note that x/ (1+ x ) ≥ 1 / 2 whenever x ≥ 1. So, if the sequence ( a n ) has infinitely many terms greater than or equal to 1, then the sequence (...
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 Spring '09
 lim, #, 2m, Sn

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