sole8 - n X k =1 | a k b k | ≤ n X k =1 a 2 k 1 2 n X k...

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ANALYSIS EXERCISE 8–SOLUTIONS 1. First, using the inequality 1 2 ( a + b ) ab , which is valid for all non-negative a and b we see that a n p for all n 2. So the sequence is bounded below. Next, we prove that it is decreasing. Indeed, a n +1 - a n = 1 2 p a n - a n = p - a 2 n 2 a n 0 . This implies that the sequence converges. Let l = lim n →∞ a n . Passing to the limit in the recurrence relation we obtain the equality l = 1 2 l + p l , which leeds to l 2 = p or l = p . 2. (a) s n = n X k =1 1 (3 k - 1)(3 k + 2) = 1 3 n X k =1 1 3 k - 1 - 1 3 k + 2 = 1 3 1 2 - 1 3 n + 2 . Therefore lim n →∞ s n = 1 6 . (b) s n = n + 2 - n + 1 + 1 - 2 = 1 n + 2 + n + 1 + 1 - 2. Therefore lim n →∞ s n = 1 - 2. 3. (a) Diverges since lim n →∞ n 2 n - 1 = 1 2 6 = 0. (b) Divereges since 1 n 1 n . (c) Diverges since lim n →∞ ( - 1) n n n +1 6 = 0. (d) Converges since 1 n n +1 1 n 3 / 2 . (e) Converges since 3 n n 5 n ( 3 5 ) n . 4. (a) From the convergence of X n =1 a n it follows that lim n →∞ a n = 0. Therefore ( N N )( n N ) [( n > N ) ( a n < 1)] . Hence for these n ’s a 2 n < a n . Apply the comparison test. (b) The assertion follows from the definition of the convergence of series and from the inequality
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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 right-hand side converges as it is the sum of con-vergent series. The left-hand 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 √ n-1) n . Therefore lim n →∞ n √ a n = lim n →∞ ( n √ n-1) = 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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