Cauchy Product of $m$ series 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?
 A: 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 collection 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.
