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We need two further constants let α = ∑ ∞ n =0 |

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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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We need two further constants Let α = ∑ ∞ n =0 | a n |...

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