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This question is motivated by the answer With $y_n$ a sequence of real numbers, prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges, where essentially the following fact is used:

Let $A = \sum_{n=0}^\infty a_n$ be an absolutely convergent series, and $(b_n)$ a convergent sequence, $b_n \to B$. Then $$ \lim_{n \to \infty} \sum_{k=0}^n a_k b_{n-k} = A B \, . $$

This is not too difficult to prove (sketch): Write $$ \sum_{k=0}^n a_k b_{n-k} = B \sum_{k=0}^n a_k + \sum_{k=0}^n a_k \bigl( b_{n-k} - B \bigr) $$ The first sum converges to $AB$. For $\varepsilon > 0$, split the second sum into two parts $$ \sum_{k=0}^{n-N} a_k \bigl( b_{n-k} - B \bigr) + \sum_{k=n-N+1}^n a_k \bigl( b_{n-k} - B \bigr) \\ = \sum_{k=0}^{n-N} a_k \bigl( b_{n-k} - B \bigr) + \sum_{j=0}^{N-1} a_{n-j} \bigl( b_j - B \bigr) $$ where $N$ is chosen such that $\lvert b_n - B \rvert < \varepsilon$ for $n \ge N$. The first part can be estimated by $\varepsilon \sum_{n=0}^\infty |a_n|$, and the second (finite) sum converges to zero.

Now I am fairly sure that this is not new and must have been done before. However, I could not find a reference. So my question is: Is there a name for the above statement, or is there some "well-known theorem" for which this is just a special case?

The term $ \sum_{k=0}^n a_k b_{n-k}$ reminds me of the Cauchy product, but nothing is given about $\sum b_n$ here. Or is it perhaps related to summation methods for series?

(Or is it so trivial that everybody just knows it?)

Martin R
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  • I don't recognise the result, so it probably doesn't have a well-known name. I think the closest you get to an actual name might be something like Exercise 2.5.3 or Example 3.5 in some specified analysis or calculus textbook somewhere. It's definitely not "so trivial", and several results that are trivial in hindsight do have their own name. I just don't think this result is important enough for that. – Arthur Jun 12 '16 at 17:52
  • I've only ever seen this result addressed as the "reverse sum product function." – Steven Harding Jun 12 '16 at 18:07

2 Answers2

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In electronics, we call such a sum : the numerical convolution of two signals. And like you've remarked, it is a Cauchy product.

This theorem was proved by Franz Mertens. What's missing from the above proof, is supposing that at least one of the series has to be absolutely convergent. Otherwise, the estimation of the first part of the second sum, isn't justified. A counterexample is given in the Wikipedia link.

ielyamani
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  • Is that really the same theorem? My question is about a series $\sum_{n=0}^\infty a_n$ and a sequence $(b_n)$, not another series. – Martin R Dec 06 '18 at 12:11
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As it turns out, the statement is equivalent to Merten's theorem about the Cauchy product of two series (with the first series being absolutely convergent).

We define $$ \begin{align} \beta_0 &= b_0 \\ \beta_n &= b_n - b_{n-1} \text{ for } n \ge 1 \, . \end{align} $$ and $\gamma_n = \sum_{k=0}^n a_k \beta_{n-k}$.

Then $b_n = \beta_0 + \ldots + \beta_n$ and $B = \sum_{n=0}^\infty \beta_n$. A simple calculation shows that $$ \sum_{k=0}^n a_k b_{n-k} = \sum_{k=0}^n \gamma_k \, . $$

Therefore $\lim_{n \to \infty} \sum_{k=0}^n a_k b_{n-k} = A B $ is equivalent to $ \sum_{n=0}^\infty \gamma_n = A B$, and that is precisely he contents of Merten's theorem.

Martin R
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  • It appears that @ielyamani answered the question nearly one year prior to your post in which he/she wrote "This theorem was proved by Franz Mertens" Given the equivalence of Merten's theorem and the theorem of interest, do you not believe that it is more appropriate to accept that originally posted answer rather than accept your own? ;-) – Mark Viola Jun 26 '21 at 16:50
  • @MarkViola: My question was about $\lim_{n \to \infty} \sum_{k=0}^n a_k b_{n-k}$ where $\sum_{n=0}^\infty a_n$ is a series and $(b_n)$ is a sequence. That is not a Cauchy product of two series. (I mentioned in the question that it reminded me of a Cauchy product, but could not see the exact connection at that time. Nor could anybody else in the next three years.) ielyamani answered “this was proved by Mertens” without pointing out the equivalence. They even deleted their answer after I asked if it is really the same. – Martin R Jun 26 '21 at 17:03
  • (Cont.) Later I realized that my problem can be transformed into a Cauchy product, and then it turns out to be equivalent to Mertens theorem. I encouraged ielyamani to undelete their answer and upvoted it. But I accepted my answer because it demonstrates that the two things are equivalent. – Martin R Jun 26 '21 at 17:03
  • So the reason you accepted yours was due to your demonstrating the equivalence while evidently ielyamani was unaware of this fact. Is that correct? – Mark Viola Jun 26 '21 at 17:06
  • @MarkViola: ielyamani answered “This is a Cauchy product, and it was proven by Mertens” which is not correct because my question was not about a Cauchy product. If they had said “this can be proved using Mertens theorem” my actions might have been different. But they deleted their answer instead. I accepted my answer because (in my opinion) it is the one which answers my question. – Martin R Jun 26 '21 at 17:12