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Unformatted text preview: n X k =1  a k b k  ≤ n X k =1 a 2 k ! 1 / 2 n X k =1 b 2 k ! 1 / 2 . 1 2 ANALYSIS EXERCISE 8–SOLUTIONS There are several other proofs as well. For example, the result also follows from the simple inequality  a k b k  ≤ (1 / 2) a 2 k + (1 / 2) b 2 k . The sum of the righthand side converges as it is the sum of convergent series. The lefthand side then converges by comparison. 5. (a) a n = 100 n n ! . Therefore lim n →∞ n √ a n = lim n →∞ 100 n √ n ! = 0 . By the Cauchy test the series converges. (b) a n = ( n √ n1) n . Therefore lim n →∞ n √ a n = lim n →∞ ( n √ n1) = 0 . By the Cauchy test the series converges. (c) a n = n 2 n . a n +1 a n = 1 2 n + 1 n . Hence lim n →∞ a n +1 a n = 1 2 . By the d’Alembert test the series converges....
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 Winter '10
 Spyros
 1 K, 3k, 2 L, 1 3K

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