How to interchange the two summations and simplify?

Question

How can we simplify the following summation? $$\sum_{i=0}^{r-j}\sum_{l=0}^{i+j-1}(-1)^{i-1}\binom{i+j}{i}\binom{r-i-1}{r-j-i}F(r,j,l).$$ Here, $F(r,j,l)$ is independent of $i$. So, how are the limits change when interchanging the summations?

Answer 1

Here is a starter. Let $a_{i,j,r}=(-1)^{i-1}\binom{i+j}{i}\binom{r-i-1}{r-j-i}$.

We obtain \begin{align*} \color{blue}{\sum_{i=0}^{r-j}}&\color{blue}{\sum_{l=0}^{i+j-1}a_{i,j,r}F_{r,j,l}}\\ &=\sum_{i=j}^r\sum_{l=0}^{i-1}a_{i-j,j,r}F_{r,j,l}\tag{1}\\ &=\sum_{{0\leq l<i\leq r}\atop{i\geq j}}a_{i-j,j,r}F_{r,j,l}\tag{2}\\ &=\sum_{l=0}^{r-1}\sum_{i=\max\{j,l+1\}}^ra_{i-j,k,1}F_{r,j,l}\tag{3}\\ &\,\,\color{blue}{=\left(\sum_{l=0}^{j-1}F_{r,j,l}\right)\left(\sum_{i=j}^{r}a_{i-j,j,r}\right)} \color{blue}{+\sum_{l=j}^{r-1}F_{r,j,l}\sum_{i=l+1}^ra_{i-j,j,r}}\tag{4} \end{align*}

Comment:

• In (1) we shift the index $i$ of the outer sum to start with $i=j$.

• In (2) we just write the index region more conveniently without changing anything else.

• In (3) we exchange the sums.

• In (4) we separate the double sum into two parts, noting that the first term is a product of single sums.