Double summation index problem I often meet the following situation:

$$\sum\limits_{n=0} ^\infty \sum\limits_{k=0} ^n f(k)g(n-k)=\sum\limits_{p=0} ^\infty \sum\limits_{q=0}^\infty f(p)g(q)$$

While intuitively this is very clear to me, I'm having problems to rigorously prove this. Could somebody please help me out?

So far my friends and I have come up with the map: $(n,k) \rightarrow (q,p) \text{ with } k\leq n$ where $q(n)=n-k$ and $p(n)=q+n=k$ Therefore we only need to prove  bijection.

Thanks

EDIT: The original problem was motivated by:

$$e^{(L_A+L_B)}=\sum\limits_{n=0} ^\infty \sum\limits_{k=0} ^n \frac{1}{k! (n-k)!} L_1^k  L_2^{n-k}=\sum\limits_{p=0} ^\infty \sum\limits_{q=0}^\infty \frac{1}{p! q!} L_1^p L_2^q=$$

where $L_A$ and $L_B \in \mathcal{B}$ and $\mathcal{B}$ is a Banach space. However, I encountered the same problem when dealing with double Fourier transforms
It seems that it is not the convergence that bothers you, but the description of the same set in two different ways. Draw a figure of ${\mathbb N}_{\geq0}\times {\mathbb N}_{\geq0}$, and everything becomes obvious. Anyway, here is a formal proof (all letters denote nonnegative integers):

For given $n\geq0$ one has
$$\{(k,n-k)\&gt;|\&gt;0\leq k\leq n\}=\{(p,q)\&gt;|\&gt;p\geq0, \ q\geq 0, \ p+q=n\}\ .$$
It follows that
$$\eqalign{\{(k,n-k)\&gt;|\&gt;n\geq0, \ 0\leq k\leq n\}
&amp;=\bigcup_{n\geq0}\{(p,q)\&gt;|\&gt;p\geq0, \ q\geq 0, \ p+q=n\}\cr
&amp;=\{(p,q)\&gt;|\&gt;p\geq0, \ q\geq 0\}\ .\cr}$$
Let us consider the term $f(p)g(q)$, where $p$, $q$ are fixed. The summation on the left-hand side contains the term $f(p)g(q)$ ($k=p$, and $n=p+q$). Moreover, the term appears only once in this summation. This implies the equality.