I would like to show that
$ \sum_{N=0}^\infty \sum_{n=0}^N f\left(n, N-n\right)$
is equal to
$ \sum_{n=0}^\infty \sum_{N-n=0}^\infty f\left(n, N-n\right).$
I can convince myself that this holds by expanding both sums to some finite (small) order and checking that every $f(x, y)$ is counted precisely once, with $x$ and $y$ varying independently.
Is there a more general or concise way of proving the equivalence of these two sums than just expanding it out and somehow "seeing" that they count every $f$ once and only once?
False.
Let $f(n,k)=1$ if $k\neq0$ and $f(n,0)=-n$.
The first sum converges, since the inner sum is always $0$.
The second does not converge, as the inner sum always diverges.