And that requires a different approach what i will

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And that requires a different approach. What I will prove here is: If n =0 a n , n =0 b n converge absolutely, then defining c n = n k =0 a k b n - k for n = 0 , 1 , 2 , . . . , thhe series n =0 c n converges absolutely and X n =0 c n = ˆ X n =0 a n ! ˆ X n =0 b n ! . The absolute convergence part is easy. One simply verifies N X n =0 | c n | = | a 0 b 0 | + | a 0 b 1 + a 1 b 0 | + · · · + | a 0 b N + · · · + a N b 0 | | a 0 | | b 0 | + | a 0 | | b 1 | + | a 1 | | b 0 | + · · · + | a 0 | | b N + · · · + | a N | | b 0 | . The last expressions is a sum of terms | a j || b k | in which every pair ( j, k ) appears at most once, and 0 j, k N . Thus the expression is less than
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or equal the sum of all possible such pairs; that is, N X n =0 | c n | | a 0 || b 0 | + | a 0 || b 1 | + · · · + | a 0 || b N | + | a 1 || b 0 | + · · · + | a 1 || b N | + cdots + | a N || b N | = ˆ N X n =0 | a n | ! ˆ N X n =0 | b n | ! ˆ X n =0 | a n | ! ˆ X n =0 | b n | ! . It follows that the partial sums of the series n =0 | c n | are uniformly bounded, hence it converges. But to what it converges has still to be established. To do this, let us set s n = n X k =0 a k , t n = n X k =0 b k , u n = n X k =0 c k , A = X k =0 a n , B = X k =0 b n . 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
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  • Spring '11
  • Speinklo
  • CN, k=0, ak sn−k

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