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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