How to formally show that following summation is equivalent?

Question

In on of the book I read that $$\sum_{i=1}^{2k+1}\sum_{m+n=i+1}Y_{mn}t^{i-1}=\sum_{m,n}^{k+1}Y_{mn}t^{m+n-2}$$ where $Y_{mn}$ is the $(m,n)th$ entry of symmetric matrix (of size $k+1$) and $t$ is some constant value from the real set. Here is my one step that can do. I take the l.h.s. and I can write $$\sum_{i=1}^{2k+1}\sum_{m+n=i+1}Y_{mn}t^{i-1}=\sum_{i=1}^{2k+1}\sum_{m=1}^{k+1}\sum_{n=i+1-m}Y_{mn}t^{i-1}\\ =\sum_{i=1}^{2k+1}\sum_{m=1}^{k+1}\sum_{n=1}^{k+1}Y_{mn}\delta(n-i-1+m)t^{i-1}$$ where $\delta(i)=1$ when $i=0$ and $0$ otherwise. What should be the next step. I guess that I should bring the summation over $i$ to inside and then I get what I want. But is it a valid step? I ask this because if we have $t=\infty$ then the summation may not converge. I think while changing the limits of integration it is required that the integral must converge but I do not know about the summation. Please clarify my confusion. Thanks in advance.

Answer 1

You are already done.

"Infinity" is not a number and it plays not role when dealing with sums (are you worried that $t+t$ "does not converge" when $t=\infty$?). The fact that you can exchange the sums is simply commutativity; you have been using it since early elementary school, if not earlier.

Answer 2

It seems to me that the difficulty in making this completely "formal" is that the formula $$\sum_{i=1}^{2k+1}\sum_{m+n=i+1}Y_{mn}t^{i-1}$$ relies on facts not stated explicitly in the formula, namely that there are no non-zero values of $Y_{mn}$ for $m<1,$ $m>k+1,$ $n<1,$ or $n>k+1.$ So first of all we have to decide what is the meaning of $\sum_{m+n=i+1}$ in formal terms, and you might argue that it means we take all pairs of integers $(m,n)$ such that $1 \leq m \leq k+1,$ $1 \leq n \leq k+1,$ and $m+n=i+1,$ which we can write as $$\sum_{m+n=i+1} f(m,n,i) = \sum_{m=1}^{k+1}\sum_{n=1}^{k+1}\delta(n-i-1+m) f(m,n,i).$$

Alternatively, you could make the interpretation that the meaning of the first sum is $$\sum_{i=1}^{2k+1}\sum_{m+n=i+1}Y_{mn}t^{i-1} = \sum_{i=1}^{2k+1}\sum_{m=\max\{1,i-k\}}^{\min\{k+1,i\}} \sum_{n=i+1-m}Y_{mn}t^{i-1},$$ restricting the range of $m$ just enough to ensure that $1 \leq n \leq k+1$ when $n= i+1-m.$ If you instead put $\sum_{m=1}^{k+1}$ on the right-hand side, I would say you have to take additional steps to deal with the terms where $n <1$ or $n > k+1.$