Is there a name for this double summation identity? What is the shortest way to illustrate that it holds?

Question

Say I have the following the expression: $$\sum\limits_{j=0}^{i-1} \sum\limits_{u=0}^{j} g(u)$$

By enumeration, it is easy to see that:

in the case when $j=0$ we have $$\sum\limits_{u=0}^{j} g(u) = g(0)$$ in the case when $j=1$ we have $$\sum\limits_{u=0}^{j} g(u) = g(0) + g(1) $$ in the case when $j=n$ we have $$\sum\limits_{u=0}^{j} g(u) = g(0) + g(1) + ... g(n)$$ ... and so for any $k \in \mathbb{N}$, the number of $g(k)$ is precisely that of the number $j$ is reaching minus $k$ plus 1 i.e.,

$$\sum\limits_{j=0}^{i-1} \sum\limits_{u=0}^{j} g(u) = \sum\limits_{j=0}^{i-1} g(j) \cdot (i-j)$$

I am aware that this can be easily proven by induction. But is there a name such this identity? Or a one-liner proof for it? I am working on an article (it is a new proof for a known theorem) and a part of it involves using this identity. So I am wondering how should I demonstrate that it holds in the simplest way possible (or is this a trial fact)?

Answer 1

It hasn´t a name, but in my opinion there is only a version of Fubinni-Cantelli's Theorem.

For a "linear proof" we have: First, for simplicity, take $n=i-1$. Thus, $\sum\limits_{j=0}^{i-1} \sum\limits_{u=0}^{j} g(u)=\sum\limits_{j=0}^{n} \sum\limits_{u=0}^{j} g(u)$.

Now, since $0\le j\le n$ and $0\le u\le j$, we can "iterated" the sum as putting $0\le u\le j\le n$ (such that in multiple integration) to obtain $\sum\limits_{j=0}^{n} \sum\limits_{u=0}^{j} g(u)=\sum\limits_{u=0}^{n} \sum\limits_{j=u}^{n} g(u)=\sum\limits_{u=0}^{n}g(u) \sum\limits_{j=u}^{n}1=\sum\limits_{u=0}^{n}g(u)(n-u+1)$.

Finally, since $n=i-1$ we have $$\sum\limits_{u=0}^{n}g(u)(n-u+1)=\sum\limits_{u=0}^{i-1}g(u)(i-u)$$

as we desired.

Answer 2

It’s basically just reversing the order of summation:

$$\begin{align*} \sum_{j=0}^{i-1}\sum_{u=0}^jg(u)&=\sum_{0\le u\le j\le i-1}g(u)\\ &=\sum_{u=0}^{i-1}\sum_{j=u}^{i-1}g(u)\\ &=\sum_{u=0}^{i-1}\big((i-1)-(u-1)\big)g(u)\\ &=\sum_{u=0}^{i-1}(i-u)g(u)\;. \end{align*}$$

Normally I’d shorten this by two steps that I’ve included here for extra clarity: I’d omit the version with the single summation, and I’d omit the penultimate line, making it a two-step calculation.