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I'm trying to understand the change of variable for this double summation:

$$\sum_{n=0}^{\infty} \sum_{m=0}^{n} \xi(m,n)=\sum_{\ell=0}^{\infty} \sum_{m=0}^{[\frac{\ell}{2}]} \xi(m,n=\ell-m)$$

where $\ell=m+n$ and $[\frac{\ell}{2}]$ means summing up to the largest integer that is less or equal than $\frac{\ell}{2}.$ So based on this question Double summation, index change clarification. I was trying to use Iverson bracket notation but I'm not sure of my deduction. What I have so far is

$$\begin{align}\sum_{n=0}^{\infty} \sum_{m=0}^{n} \xi(m,n) &= \sum_{n,m} [0\leq n<\infty]\cdot [0\leq m \leq n] \xi(m,n) \\ &= \sum_{n,m} [0\leq m \leq n < \infty] \xi(m,n) \end{align}$$

From here I could write

$$\sum_{n,m} [0\leq m <\infty]\cdot [m\leq n < \infty] \xi(m,n) = \sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \xi(m,n)$$

So to get my result I'm not sure if I can also write

$$\begin{align}[0\leq m \leq n < \infty] &\rightarrow [0\leq 2m \leq n + m < \infty] \\ &\rightarrow [0\leq 2m \leq \ell < \infty]\cdot [\ell=m+n] \\ &\rightarrow [0\leq m \leq \frac{\ell}{2}]\cdot[0 \leq \ell < \infty]\cdot [\ell=m+n]\end{align}$$

What also confuses me is if in the last line I should write the first bracket as either $2m\leq\ell$ or $2m < \infty$ to separate the conditions.

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    $\begingroup$ You could begin by simplifying some of your expressions in the Iverson brackets. $n,m,l$ are integers, of course they are less than infinity, so write $[0 \le n]$ instead of $[0 \le n < \infty]$ $\endgroup$ Commented Sep 19, 2020 at 12:22

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The work with Iverson brackets looks (besides notational aspects) fine. Note that in the derivation we not only have to show implications $\rightarrow$ but equivalence relations of the index regions we want to sum up.

Here is a derivation based upon index transformations without using Iverson brackets.

We obtain \begin{align*} \color{blue}{\sum_{n=0}^{\infty}\sum_{m=0}^n\xi(m,n)} &=\sum_{0\leq m\leq n<\infty}\xi(m,n)\tag{1}\\ &=\sum_{{0\leq m}\atop{2m\leq m+n<\infty}}\xi(m,n)\tag{2}\\ &=\sum_{{0\leq m}\atop{2m\leq l<\infty}}\xi(m,l-m)\tag{3}\\ &=\sum_{0\leq l<\infty}\sum_{0\leq 2m\leq l}\xi(m,l-m)\tag{4}\\ &\,\,\color{blue}{=\sum_{l=0}^\infty\sum_{m=0}^{\lfloor l/2\rfloor}\xi(m,l-m)} \end{align*} and the claim follows.

Comment:

  • In (1) we write the index region somewhat more conveniently.

  • In (2) we do an equivalence transformation of the inequality chain by adding $m$ as preparation for the next step.

  • In (3) we introduce $l$ and substitute $l=m+n$.

  • In (4) we rewrite the index region, so that summing over $l$ becomes the outer series.

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