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, Natural number, Universal quantification, positive integer

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