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I'm reading Complex Analysis by Freitag and Busam (2nd edition), and in exercise 14 for section I.2, they want us to prove the following variant of Abel's Test:

Given two sequences of complex numbers $(a_n)$ and $(b_n)$, if $$\sum_{n=0}^{\infty}a_n(b_n-b_{n+1})$$converges and $$(a_nb_{n+1})$$converges, then show $$\sum_{n=0}^{\infty}a_nb_n$$converges.

(I've rephrased it slightly because the actual problem makes use of the previous exercise (13) which was Abel's summation by parts, but all the given assumptions are there) My question is this: Is this actually correct? It seems phrased differently from what I find here, as there is no mention of monotone. Furthermore, I think the following is a counterexample: $$a_n=1, b_n=1+\frac{1}{n+1}$$for all $n$.

I couldn't find an errata for this text. If I misunderstood something, I hope someone here can point it out. Thank you.

Sidenote: This is not a homework problem. I'm just trying to teach myself complex analysis since my master's program never offered it (I'm slightly bitter).

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    $\begingroup$ Looked up the question and i think it's a typo and they mean $(A_nb_{n+1})$ convergent (where $A_n =\sum_{k \le n}a_k$) $\endgroup$
    – Conrad
    Dec 28, 2020 at 0:42
  • $\begingroup$ @Conrad What about the first assumption? I can see how the result directly follows if the intended first assumption is that $\sum A_n(b_n-b_{n+1})$ converges. $\endgroup$
    – JasonM
    Dec 28, 2020 at 0:56
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    $\begingroup$ yes you are right - $a_n=1, b_n=1/n, A_n=n$ and $b_n-b_{n+1}=\frac{1}{n(n+1)}$ is a counterxample, so indeed you need $\sum A_n(b_n-b_{n+1})$ convergent too so there was a double typo i guess; what i would suggest though is try and remember the technique of partial summation which is extremely important rather than the result which is a consequence of the technique $\endgroup$
    – Conrad
    Dec 28, 2020 at 1:00
  • $\begingroup$ @Conrad Thanks for the advice. The text didn't mention that exercise 13 was an analog to integration by parts, and it was written in a way that made it harder for me to see it. I only realized the analogy when I looked it up on wikipedia, and I certainly won't forget it now. $\endgroup$
    – JasonM
    Dec 28, 2020 at 1:25

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Thanks to @Conrad for verifying my counterexample and providing a likely edit to the text. They suggested more reasonable assumptions would be $\sum A_n(b_n-b_{n+1})$ and $(A_nb_{n+1})$ converge, where $$A_n=\sum_{k=0}^na_k.$$With these assumptions, the result follows immediately from the Cauchy Criteria for convergent sums and the identity $$\sum_{k=m}^n a_kb_k=A_nb_{n+1}-A_{m-1}b_m-\sum_{k=m}^nA_k(b_{k+1}-b_k)$$i.e. summation by parts, which was proven in the previous exercise.

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