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Unformatted text preview: 18.781, Fall 2007 Problem Set 11 Solutions to Selected Problems Problem 1 (a) It is very easy to find the sequence which satisfies given condition. For example, a n := 1 n (Note that this sequence has no zero term.) Then P N = Q N n =1 a n = 1 n ! , and the limit of P n is clearly 0. (b) Write as following ; Y n =1 a n = Y n =1 (1 + ( a n 1)) . Let b n := a n 1, then to show lim n b n = 0 is equivalent to show that lim n a n = 1. For the convergent infinite product, by our definition, lim n P n = where 6 = 0. Then lim n a n = lim n P n P n 1 = lim n P n lim n P n 1 = = 1 , as desired. (c) Think of b n = 1 n . Then 1 + b n = n +1 n , and we have P N = N Y n =1 n + 1 n = 2 1 3 2 N + 1 N = N + 1 . As N , P N , so it does not converge. (d) When a n > 0 for all n , we can say that log( P N ) = N n =1 log a n . Hence, [ P n converges nonzero number] is equivalent to [ n =1 log a n converges]. Therefore it is enough to show that X p log([1 p s ] 1 ) = X p log p s p s 1 = X p log 1 + 1 p s 1 < . To show that, I claim that If x > 0, then log(1 + x ) x . 1 It is not hard to prove this. For example, let f ( x ) = x log(1 + x ). Then f (0) = 0 and f ( x ) = 1 1 1+ x = x 1+ x < 0, hence f is decreasing in [0 , ], so 0 = f (0) f ( x ), and we get the conclusion....
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This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.
 Spring '09
 BRUBAKER
 Number Theory

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