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Unformatted text preview: To prove the converse, assume now only that { a nA } n =1 converges to 0. Then given any > 0, there exists a positive integer N such that  ( a nA ) < for all n N . This just says that  a nA  < for all n N . Therefore { a n } n =1 converges to A . P.55, #9. Let > 0. Since { a n } n =1 converges to A , there exists a positive integer N a such that  a nA  < for all n N a . Moreover, since { b n } n =1 converges to A , there exists a positive integer N b such that  b nA  < for all n N b . Set N = max { N a , N b } . Then we have both  a nA  < and  b nA  < for all n N . Since a n c n b n for all n , we thus have A < a n c n b n < A + for all n N , and so  c nA  < for all n N . Therefore the sequence { c n } n =1 converges to A ....
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 Fall '08
 Akhmedov,A
 Math

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