Confusion in proof regarding the summation operator

Question

Why does $\sum_{n=0}^\infty\left[\sum_{m=0}^n\left[\frac{1}{m!(n-m)!} \cdot A^{n-m} \cdot B^m\right]\right]=\sum_{l=0}^\infty\left[\frac{1}{l!} \cdot A^l\right] \cdot \sum_{m=0}^\infty\left[\frac{1}{m!} \cdot B^m\right]$ hold?

What properties are used here?

In my textbook there is a hint to perform a change-of variables $n=m+l$ and to interchange the summations.

Then we would get: $\sum_{n=0}^\infty\left[\sum_{m=0}^n\left[\frac{1}{m!(n-m)!} \cdot A^{n-m} \cdot B^m\right]\right]=\sum_{m=0}^{m+l}\left[\sum_{m+l=0}^\infty\left[\frac{1}{m!l!} \cdot A^l \cdot B^m\right]\right]$ but now I don't know how to continue, i.e. how to arrive at $\sum_{l=0}^\infty\left[\frac{1}{l!} \cdot A^l\right] \cdot \sum_{m=0}^\infty\left[\frac{1}{m!} \cdot B^m\right]$.

The summand looks fine but how to adapt the summation bounds and why?

Answer 1

We obtain \begin{align*} \color{blue}{\sum_{n=0}^\infty}&\color{blue}{\sum_{m=0}^n\frac{1}{m!(n-m)!}A^{n-m}B^m}\\ &=\sum_{0\leq m\leq n<\infty}\frac{1}{m!(n-m)!}A^{n-m}B^m\tag{1}\\ &=\sum_{m=0}^\infty\sum_{n=m}^\infty\frac{1}{m!(n-m)!}A^{n-m}B^m\tag{2}\\ &=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{m!n!}A^{n}B^m\tag{3}\\ &\,\,\color{blue}{=\left(\sum_{n=0}^\infty\frac{1}{n!}A^{n}\right)\left(\sum_{m=0}^\infty\frac{1}{m!}B^m\right)}\tag{4} \end{align*} and the claim follows.

Comment:

• In (1) we rewrite the index range as preparation for the next step.

• In (2) we exchange the order of summation provided the series converge absolutely, so that reordering does not change the value of the double series.

• In (3) we change the index of the inner series to start with $n=0$.

• In (4) we finally write the double series as product of two series.