Double sum identity

Question

Consider the following double sum identities

$$\sum_{n=0}^\infty\sum_{m=0}^n a(m,n-m) = \sum_{p=0}^\infty\sum_{q=0}^\infty a(p,q) = \sum_{r=0}^\infty\sum_{s=0}^{\lfloor{r/2}\rfloor} a(s,r-2s)$$

The first identity I understand. The relation between $(n,m)$ and $(p,q)$ is such that for any pair of naturals there is a single pair in the first range. So all pairs of naturals appear after all.

Now how do we prove the last identity? Why can we rewrite it like that, with the floor function?

Answer 1

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, $\sum_{\alpha \in A}x_\alpha = \sum_{\beta \in B}y_\beta$.

Answer 2

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.