hw9-201141 - k l =1 b l .Then we have for all N...

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Homework 9. (1) Let c n > 0 in R . Prove that lim c n +1 c n lim n c n lim n c n lim c n +1 c n . In particular, if lim n →∞ c n +1 c n exists, then lim n →∞ n c n = lim n →∞ c n +1 c n . (2) Let a n 0 and b n 0. Assume that both ( a n ) and ( b n ) are bounded sequences. (a) Prove that lim a n b n ( lim a n )( lim b n ). (b) Prove that lim a n b n = ( lim a n )( lim b n ), in case ( a n ) converges. (c) Give an example that we can have strict inequality in (a). (d) Prove lim a 2 n = ( lim a n ) 2 . (3) If n =0 a n z n has radius of convergence R , what is the radius of convergence of n =0 a n (2 z ) n and of n =0 a 2 n z n . (4) (Summation by parts) Let ( a n ) N n =1 and ( b n ) N n =1 be finite sequences of complex numbers and let B k =
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Unformatted text preview: k l =1 b l .Then we have for all N > M > 1 N X n = M a n b n = a N B N-a M B M-1-N-1 X n = M ( a n +1-a n ) B n . (Hint: Substitute b n = B n-B n-1 in the sum on the left.) (5) Prove that n =1 z n n converges for all z with | z | = 1 except z = 1. (Hint: Use the previous problem.) (6) Show that Log(1-z ) = X n =1-z n n for | z | < 1, by proving that both sides are holomorphic on | z | < 1, agree at z = 0, and have the same derivative on | z | < 1. 1...
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