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# N = 0 1 2 and that requires a different approach what

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Unformatted text preview: n = 0 , 1 , 2 ,... . 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 b | + | a b 1 + a 1 b | + ··· + | a b N + ··· + a N b | ≤ | a || b | + | a || b 1 | + | a 1 || b | + ··· + | a || b N + ··· + | a N || b | . 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 or equal the sum of all possible such pairs; that is, N X n =0 | c n | ≤ | a || b | + | a || b 1 | + ··· + | a || b N | + | a 1 || b | + ··· + | 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 ....
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n = 0 1 2 And that requires a different approach What I...

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