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Unformatted text preview: We need two further constants. Let α = ∑ ∞ n =0  a n  . Let C ≥ 0 be such that  t n  ≤ C for all n . Such a C exists because the series ∑ ∞ n =0 b n converges. (Convergent sequences are bounded.) Since B is the limit of the sequence { t n } , we also have  B  ≤ C . Let ² > 0 be given. Because ∑ ∞ n =0  a n  < ² , there is N such that ∞ X n = N +1  a n  < ² 3 C . Then also fl fl fl fl fl n X k =1 a k A fl fl fl fl fl = fl fl fl fl fl ∞ X k = n +1 a k fl fl fl fl fl ≤ ∞ X k = n +1  a k  ≤ ∞ X k = N +1  a k  < ² 3 C if n ≥ N . Because ∑ ∞ n =0 b n = B , there is M such that if n ≥ M , then  B t n  < ²/ (3 α ). Now one sees that if n > N + M n X k =0 c k = a ( b + ··· + b n ) + a 1 ( b + ··· + b n 1 ) + ··· + a n 1 ( b + b 1 ) + a n b n = n X k =0 a k s n k . If k ≤ n M then n k ≥ M . For k = 0 ,...,n M we write s n k = B + ( s n k B ) and  s n k B  < ² . Now n M > N , so ∑ n k = n M +1  a k  < ² . 2 Thus fl fl fl fl fl n X k =0 c k AB fl fl fl fl fl = fl fl fl fl fl n X k =0 a k s n k AB fl fl fl fl fl = fl fl fl fl fl n M X k =0 a k s n k AB + n X k = n M +1 a k s n k fl fl fl fl fl ≤ fl fl fl fl fl n M X k =0 a k s n k AB fl fl fl fl fl + n X k = n M +1  a k  s n k  ≤ fl fl fl fl fl n M X k =0 a k ( B + s n k B ) AB fl fl fl fl fl + n X k = n M +1  a k  s n k  = fl fl fl fl fl ˆ n M X k =0 a k A !...
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 Spring '11
 Speinklo
 CN, k=0, ak sn−k

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