Exercise 2.8.7

Assume that i = 1 a i converges absolutely to A , and j = 1 b j converges absolutely to B .

(a)
Show that the iterated sum i = 1 j = 1 | a i b j | converges so that we may apply Theorem 2.8.1.
(b)
Let s nn = i = 1 n j = 1 n a i b j , and prove that lim n s nn = AB . Conclude that
i = 1 j = 1 a i b j = j = 1 i = 1 a i b j = k = 2 d k = AB ,

where, as before, d k = a 1 b k 1 + a 2 b k 2 + + a k 1 b 1 .

Answers

(a)
Let i = 1 | a i | converge to A and j = 1 | b j | converge to B . By the Algebraic Limit Theorem for Series,
i = 1 j = 1 | a i b j | = i = 1 ( | a i | j = 1 | b j | ) = ( i = 1 | a i | ) ( j = 1 | b j | ) = A B
(b)
Theorem 2.8.1 shows that lim n s nn = i = 1 j = 1 a i b j which by the same manipulation as used in part (a), equals AB . By Theorem 2.8.1 the rest of the problem is solved.
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2022-01-27 00:00
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