Double sum identity

Considering formal series we obtain \begin{align*} \color{blue}{\sum_{r=0}^\infty\sum_{s=r}^{\left\lfloor r/2\right\rfloor} a(s,r-2s)} &=\sum_{r=0}^\infty\sum_{s=0}^r a(s,2r-2s)+\sum_{r=0}^\infty\sum_{s=0}^r a(s,2r+1-2s)\tag{1}\\ &=\sum_{0\leq s\leq r\leq \infty}\left(a(s,2(r-s))+a(s,2(r-s)+1)\right)\tag{2}\\ &=\sum_{{0\leq s\leq \infty}\atop{0\leq r-s\leq \infty,\ 0\leq r\leq \infty}}\left(a(s,2(r-s))+a(s,2(r-s)+1)\right)\tag{3}\\ &=\sum_{{0\leq s\leq \infty}\atop{0\leq q\leq \infty}}\left(a(s,2q)+a(s,2q+1)\right)\tag{4}\\ &=\sum_{{0\leq s\leq \infty}\atop{0\leq q\leq \infty}}a(s,q)\tag{5}\\ &\,\,\color{blue}{=\sum_{p=0}^\infty\sum_{q=0}^\infty a(p,q)}\tag{6} \end{align*}

Comments:

• In (1) we split the double sum with respect to even and odd $r$.

• In (2) we write the index region somewhat more conveniently and collect all terms in one double sum.

• In (3) we reformulate the index region as preparation for the next step.

• In (4) we introduce a new summation index $q = r-s$ and skip $r$.

• In (5) we simplify the expression (by applying the associative law).

• In (6) we finally replace the index $s$ with $p$ and switch back to the original index notation.

Write $K = \{(r, s) \colon r, s \in \mathbb{N}, \ r \geqslant 2s\}$. The functions: \begin{align*} & f \colon \mathbb{N}^2 \to K, \ (p, q) \mapsto (2p + q, p), \\ & g \colon K \to \mathbb{N}^2, \ (r, s) \mapsto (s, r - 2s) \end{align*} are mutually inverse bijections. Therefore, if the family $(a(p, q))_{p, q \in \mathbb{N}}$ is absolutely summable [see below]: $$ \sum_{p=0}^\infty\sum_{q=0}^\infty a(p,q) = \sum_{p, q \in \mathbb{N}}a(p, q) = \sum_{(r, s) \in K} a(s, r - 2s) = \sum_{r=0}^\infty\sum_{s=0}^{\lfloor{r/2}\rfloor} a(s,r-2s). $$ The first and third equalities follow from Dieudonne, Foundations of Modern Analysis (1969), proposition (5.3.6):

Let $(x_\alpha)_{\alpha \in A}$ be an absolutely summable family of elements of a Banach space $E$. Let $(B_n)$ be an infinite sequence of nonempty subsets of $A$, such that $A = \bigcup_nB_n$, and $B_p \cap B_q = \emptyset$ for $p \ne q$; then, if $z_n = \sum_{\alpha \in B_n}x_\alpha$ [it has already been shown that such a sum is well-defined], the series $(z_n)$ is absolutely convergent, and $$\sum_{n=0}^\infty z_n = \sum_{\alpha \in A}x_\alpha$$ ("associativity" of absolutely convergent series).

As for the second equality, quoting Dieudonne again:

Let $A$ be any denumerable set. We say that a family $(x_\alpha)_{\alpha \in A}$ of elements of a Banach space $E$ is absolutely summable if, for a bijection $\varphi$ of $\mathbb{N}$ onto $A$, the series $(x_{\varphi(n)})$ is absolutely convergent; it follows from [the rearrangement theorem for absolutely convergent series] that this property is independent of the particular bijection $\varphi$, and that we can define the sum of the family $(x_\alpha)_{\alpha \in A}$ as $\sum_{n=0}^\infty x_{\varphi(n)}$, which we write $\sum_{\alpha \in A}x_\alpha$.

Clearly, therefore, if $f \colon A \to B$ and $g \colon B \to A$ are mutually inverse bijections, and we put $y_\beta = x_{g(\beta)}$ ($\beta \in B$), so that also $x_\alpha = y_{f(\alpha)}$ ($\alpha \in A$), then the family $(x_\alpha)_{\alpha \in A}$ is absolutely summable if and only if the family $(y_\beta)_{\beta \in B}$ is absolutely summable; and in that case, $\sum_{\alpha \in A}x_\alpha = \sum_{\beta \in B}y_\beta$.