Trying to understand a double summation formula I am trying to understand a double summation mentioned in this paper: https://doi.org/10.1016/0024-3795(95)00696-6. How can we prove that this is true?
$$\sum_{l = 1}^n\;\; \sum_{k = l}^{n+l} = \sum_{k \,=\, 1}^n \;\; \sum_{l \,=\, 1}^{k} \,+\, \sum_{k \,=\, n+1}^{2n}\;\; \sum_{l \,=\, k-n}^{n}$$

Here is another variation to show the identity for $n\geq 1$:
\begin{align*}
\sum_{l=1}^n\sum_{k=l}^{n+l}a_{k,l}
=\sum_{k=1}^n\sum_{l=1}^ka_{k,l}+\sum_{k=n+1}^{2n}&\sum_{l=k-n}^na_{k,l}\tag{1}
\end{align*}
We start with the right-hand side of (1) and consider at first the left sum.

We obtain
\begin{align*}
\sum_{k=1}^n\sum_{l=1}^ka_{k,l}=\sum_{\color{blue}{1\leq l\leq k\leq n}} a_{k,l}=\sum_{l=1}^n\sum_{k=l}^{n}a_{k,l}\tag{2}
\end{align*}

In the middle of (2) we represent the index range of summation as inequality chain. This might help to better see the region of validity and how to exchange the sums.

Now the other sum. We obtain
\begin{align*}
\sum_{k=n+1}^{2n}&\sum_{l=k-n}^na_{k,l}\\
&=\sum_{k=1}^{n}\sum_{l=k}^na_{k+n,l}
=\sum_{\color{blue}{1\leq k\leq l\leq n}} a_{k+n,l}=\sum_{l=1}^n\sum_{k=1}^la_{k+n,l}\tag{3}\\
&=\sum_{l=1}^n\sum_{k=n+1}^{n+l}a_{k,l}\tag{4}
\end{align*}

Comment:

*

*In (3) we shift the index $k$ by $n$ to start with $k=1$. We are now in the same situation as in (2) and continue accordingly.


*In (4) we shift the index back.

Adding (2) and (4) gives
\begin{align*}
\sum_{l=1}^n\sum_{k=l}^{n}a_{k,l}+\sum_{l=1}^n\sum_{k=n+1}^{n+l}a_{k,l}
\color{blue}{=\sum_{l=1}^n\sum_{k=l}^{n+l}a_{k,l}}
\end{align*}
and the claim (1) follows.

