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
 A: 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
$$\{(k,n-k)\>|\>0\leq k\leq n\}=\{(p,q)\>|\>p\geq0, \ q\geq 0, \ p+q=n\}\ .$$
It follows that
$$\eqalign{\{(k,n-k)\>|\>n\geq0, \ 0\leq k\leq n\}
&=\bigcup_{n\geq0}\{(p,q)\>|\>p\geq0, \ q\geq 0, \ p+q=n\}\cr
&=\{(p,q)\>|\>p\geq0, \ q\geq 0\}\ .\cr}$$
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Completely ignoring matters of convergence, and using notations from Dijkstra et al. (see, e.g., EWD1300 for details), I would prove this equality as follows (I'm taking baby steps here):
$$\calc
\langle + n,k : 0 \leq k \leq n : f(k) \times g(n-k) \rangle
\calcop{=}{logic: introduce name $\;q\;$ using one-point rule}
\langle + n,k,q : q = n - k \land 0 \leq k \leq n : f(k) \times g(q) \rangle
\calcop{=}{arithmetic}
\langle + n,k,q : n = q - k \land 0 \leq k \leq n : f(k) \times g(q) \rangle
\calcop{=}{logic: substitute $\;n\;$ using one-point rule}
\langle + k,q : 0 \leq k \leq q - k : f(k) \times g(q) \rangle
\calcop{=}{arithmetic: simplify}
\langle + k,q : 0 \leq k \land 0 \leq q : f(k) \times g(q) \rangle
\endcalc$$
A: 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.
