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Let $\{a_n\}$, $\{b_n\}$ be sequences. Define $\displaystyle c_n=\sum_{k=1}^n a_kb_{n+1-k}$.

Prove that if $~\sum a_n=A~$ , $~\sum b_n=B~$ , and $~\sum c_n=C~$ (so they are all convergent series) then $C=AB$. (Note that we do not need $\sum a_n$ to be absolutely convergent).

Hello everyone. I am stuck on how to start this problem. I don't want the answer, just a hint on how to get started.

JRC
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James Anderson
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  • Hint: What is the general term in the product of sequences $(\sum a_{n})(\sum b_{n})$? – Representation Dec 07 '20 at 09:10
  • @MartinR I looked at the question you reference. I have to admit that the answer is not obvious to read on my side... Yes it implies that $A(x)B=C(x)$ for $0\lt x \lt 1$ where $C(x)$ is the Cauchy product of $A(x)$ and $B$. Yes also $\lim\limits_{x \to 1^-} C(x)$ exists as $\lim\limits_{x \to 1^-} A(x)=A$. What I don't see is why $\lim\limits_{x \to 1^-} C(x)=C$? Which is in fact the question raised. – mathcounterexamples.net Dec 07 '20 at 09:30
  • @mathcounterexamples.net: That follows from Abel's theorem, as I understand it. The convergence of $C= \sum c_n$ implies that $\lim_{x \to 1_-} C(x) = C$. The same applies to the other two series. – Martin R Dec 07 '20 at 09:33
  • It is clear that $\lim\limits_{x \to 1^-} A(x)=A$ is a consequence of Abel's theorem. But $C(x)$ has a form that is not the one of Abel's theorem, i.e. something like $\sum_n c_n x^n$. It is $\sum_n (\sum_k a_k b_{n-k} x^k)$. – mathcounterexamples.net Dec 07 '20 at 09:37
  • @mathcounterexamples.net: I think it has. With $A(x) = \sum a_n x^n$ and $B(x) = \sum b_n x^n$ you have $A(x)B(x) = C(x)$ for $|x| < 1$, where $C(x) = \sum c_n x^n$ and $c_n = \sum_{k=0}^n a_k b_{n-k}$. (I'm using zero-based indices for convenience.) This is the Cauchy product of power series. Then $x \to 1-$ and Abel's theorem gives $AB = C$. Or am I misunderstanding something? – Martin R Dec 07 '20 at 09:43
  • @MartinR You're right. I was only thinking of Cauchy product for series and not power series. Thanks! – mathcounterexamples.net Dec 07 '20 at 09:46
  • The problem is, the teacher has not taught us about Merten's theorem or Abel's theorem so I am not allowed to use those. – James Anderson Dec 09 '20 at 06:04

1 Answers1

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I'm sorry that I misunderstood the question previously. What you are looking for is probably this , which says :

Let $\sum a_{n}~$ , $\sum b_{n}$ are conditionally convergent complex series, $\sum c_{n}$ is the Cauchy product of $\sum a_n$, $\sum b_n$ such that $\sum c_n$ converges. Then, $$\sum c_{n} = \left(\sum a_{n}\right)\left(\sum b_{n}\right)$$

For a complete proof, please refer to same link as above.

EDIT : Updated the links. Sorry for inconvenience.

JRC
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    This is not an answer to the posted question. Merten's theorem supposes that one of the series is absolutely convergent. This is explicitly mentioned by the OP as not being the case. – mathcounterexamples.net Dec 07 '20 at 09:17
  • If none of the two series absolutely converges, then is the result necessarily true ? – JRC Dec 07 '20 at 09:18
  • Yes, because it is assumed in the question that $\sum c_n$ is convergent. See the possible duplicate target pointed out at the question. – Martin R Dec 07 '20 at 09:19
  • Understood. Thanks. Let me edit my answer. – JRC Dec 07 '20 at 09:26
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    Your links still point to the standard version of Mertens' theorem and its proof, which requires that one of the series is absolutely convergent. – Martin R Dec 07 '20 at 09:48
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    Sorry. I forgot to update the links. Can you please check now ? – JRC Dec 07 '20 at 11:23