Change summation order I need to change the summation order in the sum
$$
\sum_{m,n=0}^\infty  \left( \sum_{l=0}^{n+m} \left( \sum_{t=0}^l \binom{m}{t} \binom{n}{l-t}  a_{n-l+2t,m+l-2t}  \right) \right) \frac{ b_{m,n}}{m! n!}.
$$
By experiments I get the following conjecture
$$
\sum_{m,n=0}^\infty  \left( \sum_{l=0}^{n+m} \left( \sum_{t=0}^l \binom{m}{t} \binom{n}{l-t}  a_{n-l+2t,m+l-2t}  \right) \right) \frac{ b_{m,n}}{m! n!}=\sum_{t,l=0}^\infty \left( \sum_{m=0}^t \sum_{n=0}^l  \binom{t}{m} \binom{l}{n}  a_{t-m+n,l-n+m}\, b_{t+l-m-n,m+n}\right) \frac{1}{t! l!}.
$$
Any ideas how to prove it?
 A: Well, let's see.
$\begin{array}\\
\sum_{m,n=0}^\infty   \sum_{l=0}^{n+m}  \sum_{t=0}^l
&=\sum_{m=0}^\infty\sum_{n=0}^\infty   \sum_{l=0}^{n+m}  \sum_{t=0}^l\\
&=\sum_{k=0}^\infty\sum_{n=0}^k   \sum_{l=0}^{k}  \sum_{t=0}^l
\qquad k = n+m, n \le k, m = k-n\\
&=\sum_{k=0}^\infty\sum_{n=0}^k   \sum_{t=0}^{k}  \sum_{l=t}^k
\qquad t \le l \implies l \ge t\\
&=\sum_{k=0}^\infty\sum_{t=0}^{k}  \sum_{l=t}^k\sum_{n=0}^k
\qquad\text{move sums around}\\
&=\sum_{t=0}^{\infty} \sum_{k=t}^\infty \sum_{l=t}^k\sum_{n=0}^k
\qquad t \le k \implies k \ge t\\
&=\sum_{t=0}^{\infty}\sum_{l=t}^{\infty} \sum_{k=l}^\infty \sum_{n=0}^k
\qquad t \le l \le k
\end{array}
$
A: Note: This is only a partial answer. The idea of the steps below is when doing rearrangements we also take care of $\binom{p}{q}=0$ whenever we have integral $0< p<q$.

We obtain
\begin{align*}
\color{blue}{\sum_{m,n=0}^\infty}&\color{blue}{ \sum_{l=0}^{n+m} \sum_{t=0}^l \binom{m}{t} \binom{n}{l-t}  a_{n-l+2t,m+l-2t}   \frac{ b_{m,n}}{m! n!}}\\
&=\sum_{m,n=0}^{\infty}\left(\sum_{l=0}^m\sum_{t=0}^l\binom{m}{t}\binom{n}{l-t}\right.\\
&\qquad\qquad\quad\left.+\sum_{l=m+1}^{n+m}\sum_{t=0}^{\color{blue}{m}}\binom{m}{t}\binom{n}{l-t}\right)
 a_{n-l+2t,m+l-2t}   \frac{ b_{m,n}}{m! n!}\tag{1}\\
 &=\sum_{m,n=0}^{\infty}\left(\sum_{t=0}^m\sum_{l=t}^m\binom{m}{t}\binom{n}{l-t}a_{n-l+2t,m+l-2t}   \frac{ b_{m,n}}{m! n!}\right.\\
 &\qquad\qquad\quad\left.+\sum_{t=0}^m\sum_{l=1}^{n}\binom{m}{t}\binom{n}{l+m-t}\right)a_{n-m-l+2t,2m+l-2t}   \frac{ b_{m,n}}{m! n!}\tag{2}\\
 &=\sum_{m,n=0}^{\infty}\sum_{t=0}^m\binom{m}{t}\left(\sum_{l=0}^{m-t}\binom{n}{l}a_{n-l+t,m+l-t}   \frac{ b_{m,n}}{m! n!}\right.\\
 &\qquad\qquad\qquad\qquad\quad\left.+\sum_{l=1}^{n}\binom{n}{l+m-t}\right)a_{n-m-l+2t,2m+l-2t}   \frac{ b_{m,n}}{m! n!}\tag{3}\\
  &=\sum_{m,n=0}^{\infty}\sum_{t=0}^m\binom{m}{t}\left(\sum_{l=0}^{t}\binom{n}{l}a_{n+m-l-t,l+t}   \frac{ b_{m,n}}{m! n!}\right.\\
 &\qquad\qquad\qquad\qquad\quad\left.+\sum_{l=1}^{\color{blue}{n-t}}\binom{n}{l+t}\right)a_{n+m-l-2t,l+2t}   \frac{ b_{m,n}}{m! n!}\tag{4}\\
   &=\sum_{m,n=0}^{\infty}\sum_{t=0}^m\binom{m}{t}\left(\sum_{l=0}^{t}\binom{n}{l}a_{n+m-l-t,l+t}   \frac{ b_{m,n}}{m! n!}\right.\\
 &\qquad\qquad\qquad\qquad\quad\left.+\sum_{l=t+1}^{n}\binom{n}{l}\right)a_{n+m-l-t,l+t}   \frac{ b_{m,n}}{m! n!}\tag{5}\\
 &=\sum_{m,n=0}^{\infty}\sum_{t=0}^m\binom{m}{t}\sum_{l=0}^{n}\binom{n}{l}a_{n+m-l-t,l+t}   \frac{ b_{m,n}}{m! n!}\tag{6}\\
 &\,\,\color{blue}{=\sum_{t,l=0}^{\infty}\sum_{m=0}^t\binom{t}{m}\sum_{n=0}^{l}\binom{l}{n}a_{l+t-n-m,n+m}   \frac{ b_{t,l}}{t! l!}}\tag{7}\\
\end{align*}
Now we have with (7) a representation where the two inner sums  and the binomial coefficients show the same form as OPs right-hand side. Nevertheless it needs an additional twist to adjust the indices of $b_{t,l}$ which I don't see at the moment.

Comment:

*

*In (1) we split the sum using that $\binom{m}{t}=0$ if $t > m$, so that the upper index of the right-hand inner sum is set to $m$.


*In (2) we exchange the inner left-hand sums respecting the index range $0\leq t\leq l\leq m$. We also simply exchange the right-hand inner sums and we shift the index to start with $l=1$.


*In (3) we can now factor out $\sum_{t=0}^m\binom{m}{t}$ and we shift the index of the left inner sum by $t$ to start with $l=0$.


*In (4) we do a reordering of the terms by $t\to m-t$. We also set the upper index of the right-most inner sum from $n$ to $n-t$, since other values do not contribute according to $\binom{n}{l+t}=0$.


*In (5) we shift the index of the right inner sum by $t$ to start with $l=t+1$.


*In (6) we can  finally merge the two inner sums and obtain the wanted structure of the sums.


*In (7) we replace $m\leftrightarrow t$ and $n\leftrightarrow l$.
