Finite double sum: Improve index transformation In order to prove a rather complicated binomial identity a small part of it implies a transformation of a double sum. > The double sum and its transformation have the following shape: \begin{align*} \sum_{k=0}^{l}\sum_{j=\max\\{1,k\\}}^{\min\\{l,k+c\\}}1=\sum_{j=1}^{l}\sum_{k=0}^{\min\\{j,c\\}}1\qquad\qquad l\geq 1, c\geq 1 \end{align*} Here I do not want to take care of the terms which are to sum up. They are set to $1$ for the sake of simplicity. What matters is an efficient, short indextransformation showing that the identity is valid. At the time I've found a rather long-winded solution. It is added as answer to this question. But in fact I would appreciate to find a more elegant way to prove this identity.

Perhaps the simplest argument is geometric. A sketch makes it evident that if $c\le\ell$, the lefthand side is the number of points of $\Bbb N\times\Bbb N$ lying in the convex hull of the points $\langle 0,1\rangle$, $\langle 0,c\rangle$, $\langle 1,1\rangle$, $\langle \ell-c,\ell\rangle$, and $\langle\ell,\ell\rangle$. For $c\le j\le\ell$ the $j$-section has $c+1$ points, and for $1\le j
$$\sum_{k=0}^{\min\\{j,c\\}}1=\min\\{j+1,c+1\\}\;.$$

Here’s the sketch:

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The colored regions contain the points in question. The orange region comprises the $j$-sections with $c+1$ points; the blue, the $j$-sections with $j+1$ points.

For $c>\ell$ the blue region expands to the top of the square.