Check the range is the same: $\sum\limits_{\ell=0}^k\sum\limits_{j=\ell}^m=\sum\limits_{j=0}^m\sum\limits_{\ell=0}^{\min(j,k)}$? Given that $m\ge k$, why the range of $$\sum_{\ell=0}^k\sum_{j=\ell}^m$$ is the same as
$$\sum_{j=0}^m\sum_{\ell=0}^{\min(j,k)}$$
Instead of write down all to check, is there any systematical way?
I thought the latter would be $\displaystyle\sum\limits_{j=0}^m\sum\limits_{\ell=0}^{j[j\le k]+(-1)[j>k]}$, but it seems like this is wrong, can anyone help me point out way? ($\textrm{if P is True, }[P]=1, \textrm{otherwise }[P]=0$.)
 A: Hint: You can prove it, by thinking about a matrix which is $k \times m$. One side is the sum of elements at the upper part of the matrix row by row, and the other is pertinent to summing those elements column by column (For example, see this post).
A: Part 1: Solution
Algebraic way: It takes me some time to find out that $+$-operation here is combinatorically independent, so
\begin{align}\sum_{\ell=0}^k\sum_{j=l}^{m}&=\sum_{l=0}^k\left(\sum_{j=l}^k+\sum_{j=k+1}^m\right)\\
&=\sum_{j=0}^k\sum_{l=0}^j+\sum_{j=k+1}^{m}\sum_{l=0}^{k}\\
&=\sum_{j=0}^m\sum_{l=0}^{\min(j,k)}\quad\quad\quad\quad\square\tag{*}
\end{align}
For $(\textrm{*})$: since
\begin{align}
\begin{cases}j\le k & \textrm{pick } j\\j\gt k & \textrm{pick } k
\end{cases}
\implies \min(j,k).
\end{align}

Part 2: Correction
\begin{align}
\sum_{j=0}^m\sum_{l=0}^{j[j\le k]+(-1)[j\gt k]}&=\left(\sum_{j=0}^{k}+\sum_{j=k+1}^m\right)\sum_{l=0}^{j[j\le k]+(-1)[j\gt k]}\\
&=\sum_{j=0}^k\sum_{l=0}^j+\sum_{j=k+1}^m\sum_{l=0}^{-1}\\
&=\sum_{l=0}^k\sum_{j=l}^k.
\end{align}
Now it's obvious...
A: We can transform the double-sum by conveniently writing the index-regions. We obtain
\begin{align*}
\sum_{l=0}^k\sum_{j=l}^m a_{l,j}&= \sum_{\color{blue}{{0\leq l\leq k}\atop{l\leq j\leq m}}}a_{l,j}
=\sum_{j=0}^m\sum_{l=0}^{\min\{k,j\}}a_{l,j}
\end{align*}
We can easily see from the two double-inequalities in the middle of the equality chain, that $0\leq j\leq m$ and $0\leq l\leq \min\{k,j\}$.
A: For each value of $\ell \in \{0, 1, \ldots, k\}$ in the outer sum on the left hand side, the inner sum is the sum of the values of a given function, call it $f,$ on the set $A_\ell = \{(j, \ell) : \ell \leqslant j \leqslant m\}.$ The sets $A_\ell$ are pairwise disjoint, and their union is
$$
C = \{(j, \ell) \in \mathbb{N}^2 : 0 \leqslant \ell \leqslant k \text{ and } \ell \leqslant j \leqslant m\}.
$$
For each value of $j \in \{0, 1, \ldots, m\}$ in the outer sum on the right hand side, the inner sum is the sum of the values of the function $f$ on the set $B_j = \{(j, \ell) : 0 \leqslant \ell \leqslant \min\{j, k\}\}.$ The sets $B_j$ are also pairwise disjoint, and their union is
\begin{align*}
& \phantom{{}={}}
\{(j, \ell) \in \mathbb{N}^2 : 0 \leqslant j \leqslant m \text{ and } 0 \leqslant \ell \leqslant \min\{j, k\}\} \\
& = \{(j, \ell) \in \mathbb{N}^2 : 0 \leqslant j \leqslant m \text{ and } 0 \leqslant \ell \text{ and } \ell \leqslant j \text{ and } \ell \leqslant k\} \\
& = \{(j, \ell) \in \mathbb{N}^2 : \ell \leqslant j \leqslant m \text{ and } 0 \leqslant \ell \leqslant k\} \\
& = C.
\end{align*}
The iterated sums on the two sides of the claimed identity are therefore equal, because they are both equal to the sum of the function $f$ over the set $C,$ which can be written as
$$
\sum_{(j, \ell) \in C}f(j, \ell).
$$
