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Is the limit of the sequence t n we also have b c let

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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 0 ( b 0 + · · · + b n ) + a 1 ( b 0 + · · · + b n - 1 ) + · · · + a n - 1 ( b 0 + 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

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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 ! B + n - M X k =0 a k ( 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 !fl fl fl fl fl | B | + n - M X k =0 | a k | ² 3 α + C ² 3 C = fl fl fl fl fl ˆ n - M X k =0 a k - A !fl fl fl fl fl | B | + α ² 3 α + C ² 3 C < e 3 C | B | + 2 ² 3 < ² 3 + 2 ² 3 = ².
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• Spring '11
• Speinklo
• CN, k=0, ak sn−k

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