(Follow-up) Cauchy product of more than two series

Question

From Wikipedia: \begin{equation} \prod_{i=1}^m \left( \sum_{k_i = 0}^\infty a_{i,k_i} \right)=\lim_{n\to\infty}\underbrace{\sum_{k_1 = 0}^n \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}}_{=S_n} \end{equation} I find it difficult to wrap my head around the expression on the right. Can $S_n$ be rewritten in a more suggestive way? For example, does the equality \begin{equation} \forall n:S_n=\sum_{k_1+\dots+k_m\leq n}a_{1, k_1}\cdots a_{m, k_m} \end{equation} hold? Of course this different from asking if \begin{equation} \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k_1+\dots+k_m\leq n}a_{1, k_1}\cdots a_{m, k_m} \end{equation} which was my initial question.

Answer 1

First I would like to point out the abuse of notation. The letter $n$ is overloaded, since it's both the number of series and the index that tends to infinity. I'll use $m$ as the number of series.

If you carefully check the blue indices in the sum $$S_n=\sum_{k_1 = 0}^n \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_m = 0}^{k_{m-1}} a_{1, \color{blue}{k_m}} a_{2, \color{blue}{k_{m-1} - k_m}} \cdots a_{m, \color{blue}{k_1 - k_2}}$$ you'll check that their sum is $k_1$, which is from $0$ to $n$. Therefore $$S_n=\sum_{k_1+\dots+k_m\leq n}a_{1, k_1}\cdots a_{m, k_m}. $$ This form of writing this sum is propper as long as we know that we consider only integer nonnegative indices.

We can write it another way, which is closer to the idea that the Cauchy product of series is also a series. Namely, $$S_N = \sum_{n=0}^N c_n\text{ where } c_n=\sum_{k_1+\dots+k_m = n}a_{1, k_1}\cdots a_{m, k_m}.$$

To remember this idea I like thinking of multiplying the polynomials / power series. If we have the product $$\prod_{i=1}^m \left( \sum_{k_i = 0}^\infty a_{i,k_i}x^{k_i}\right)$$ then multiplying them formally we get the power series $\sum c_nx^n$ where $c_n$ is a sum of products of all coefficients $a_{i,k_i}$ such that their indices $k_i$ add up to $n$.