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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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 Spring '11
 Speinklo
 CN, k=0, ak sn−k

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