Interchanging sum over multi-indices Let $f:\mathbb{N}_0\times\mathbb{N}_0\times\mathbb{N}_0^n\to\mathbb{R}$. Suppose that
$$0=\sum_{k=0}^N\sum_{j=0}^{N-k}\sum_{|\gamma| = N-j-k}f(k,j,\gamma)$$
for every $N\in\mathbb{N}$.
I believe that
$$\sum_{|\gamma|<N}\sum_{k=0}^{N-|\gamma|-1}\sum_{j=0}^{N-|\gamma|-k-1}f(k,j,\gamma) = f(0,0,0)$$
for every $N\in\mathbb{N}$,
but I'm having trouble proving it, I think I'm messing up when interchanging the order of summations, and how the sum behaves when $|\gamma|$ is fixed. I was hoping someone could give me a hint.
Edit:
There was a mistake with the first equality, there were a few missing terms. I've managed to prove that
$$\sum_{|\gamma|<N}\sum_{k=0}^{N-|\gamma|-1}\sum_{j=0}^{N-|\gamma|-k-1}f(k,j,\gamma)= \sum_{k=0}^{N-1}\sum_{j=0}^{N-k-1}\sum_{|\gamma|<N-k-j}f(k,j,\gamma)$$
plugging in $N=1$ and $N=2$, it seems that it works, but I can't prove for an aribitrary $N$. I've tried induction, but without much success.
 A: Define $G(N)$ to be the expression you're assuming is $0$ for all $N$, i.e.
$$
G(N)=\sum_{k=0}^N\sum_{j=0}^{N-k}\sum_{|\gamma| = N-j-k}f(k,j,\gamma).
$$
In English, the assumption $G(N) = 0$ means: if we sum $f$ over all $(j,k,\gamma)$ with $j+k+|\gamma| = N$ then we'll get 0.
We can rewrite the equation by summing over all the same combinations of $(j,k,\gamma)$ but putting the $N$ sum on the outside. This might be what you want?
$$0 = \sum_{|\gamma| \le N|} \sum_{k=0}^{N-|\gamma|} f(k, N-|\gamma|-k, \gamma).$$
Note that there's no longer a sum over $j$, because once $\gamma$ and $k$ are chosen, $j$ is fully determined by the requirement $j + k + |\gamma| = N$.

Your question doesn't make it totally clear what manipulation you're hoping to see, so here's another version that might be what you want: We also know that $\sum_{n=1}^N G(n) = \sum_{n=0}^N 0 = 0$. In English, $\sum_{n=1}^N G(n)$ is the sum of $f$ over all $(j,k,\gamma)$ with $j+k+|\gamma| \le N$. (Note the original version's $=$ has now become $\le$.) Writing this in summation notation with the $\gamma$ sum on the outside gives:
$$0 = \sum_{|\gamma| \le N|} \sum_{k=0}^{N-|\gamma|} \sum_{j=0}^{N-|\gamma|-k} f(k, j, \gamma).$$
