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From Wikipedia: $$\prod_{i=1}^m \left( \sum_{k_i = 0}^\infty a_{i,k_i} \right)=\sum_{k_1 = 0}^\infty \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_m = 0}^{k_{m-1}} a_{1, k_m} a_{2, k_{m-1} - k_m} \cdots a_{m, k_1 - k_2}=:l$$ ($l$ stands for "limit" and we are looking at sequences of elements of some banach algebra). I find it difficult to wrap my head around the expression on the right.

This leads to my my question: Is the equation $$l=\sum_{n=0}^{\infty}\sum_{k_1+\dots+k_m=n}a_{1, k_1}\cdots a_{m, k_m}$$ right?

Filippo
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    To derive this formula its useful to generalize the series from a sum over $a_{j,k_j}$ to a sum over $a_{j,k_j} x^{k_j}$. The powers of $x$ acts as a book-keeping device. Try to compute what the coefficient of the $x^n$ term is when you multiply out the product. – Winther Jun 25 '20 at 11:58
  • The index $k_1 + \ldots + k_n = n$ in the last summation doesn't seem right to me... – obr Jul 12 '22 at 19:31
  • @obr Sorry, it was not a good idea to edit the question, as I changed the actual question. Thus, I undid the edit and asked the new question separately here. What question are you referring to, the old one or the new one? – Filippo Jul 12 '22 at 19:37
  • @Filippo -- Actually, both have the same issue. – obr Jul 12 '22 at 23:10
  • @obr Now I understand what you mean, thank you for pointing that out. I have adopted the notation of the accepted answer. – Filippo Jul 13 '22 at 05:30

1 Answers1

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Winther’s suggestion in the comments is a good one. Consider the following product of $m$ power series:

$$\prod_{i=1}^m\sum_{k_i\ge 0}a_{i,k_i}x^{k_i}$$

For $n\ge 0$ what is the coefficient of $x^n$? The product is

$$(a_{1,0}+a_{1,1}x+a_{1,2}x^2+\ldots)\cdot(a_{2,0}+a_{2,1}x+a_{2,2}x^2+\ldots)\cdot\ldots\cdot(a_{m,0}+a_{m,1}x+a_{m,2}x^2+\ldots)\;,$$

so each term in the product (before collecting like terms) has the form

$$a_{1,k_1}x^{k_1}\cdot a_{2,k_2}x^{k_2}\cdot\ldots\cdot a_{m,k_m}x^{k_m}=a_{1,k_1}a_{2,k_2}\ldots a_{m,k_m}x^{k_1+k_2+\ldots+k_m}\;.$$

This is an $x^n$ term iff $k_1+k_2+\ldots+k_m=n$. After collecting terms, therefore, the $x^n$ term will be

$$\sum_{\substack{k_1,\ldots,k_m\ge 0\\k_1+k_2+\ldots+k_m=n}}a_{1,k_1}a_{2,k_2}\ldots a_{m,k_m}x^n\;,$$

making the product series

$$\prod_{i=1}^m\sum_{k_i\ge 0}a_{i,k_i}x^{k_i}=\sum_{n\ge 0}\left(\sum_{\substack{k_1,\ldots,k_m\ge 0\\k_1+k_2+\ldots+k_m=n}}a_{1,k_1}a_{2,k_2}\ldots a_{m,k_m}\right)x^n\;.$$

Set $x=1$, and you have the desired result.

Brian M. Scott
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