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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?

Calum Gilhooley
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Oily
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    Note that: $$0 \leq l \leq i+j-1, ~~~~\text{and}~~~~ 0 \leq i \leq r-j.$$ Combining both inequalities, we can see that: $$0\leq l \leq i+j-1 \leq (r-j)+j-1 = r-1,$$ and from this: $$0\leq l \leq r-1 ~~~~\text{and}~~~~ l - j +1 \leq i \leq r-j.$$ Therefore, a change in order of summation would yield: $$\sum_{i=0}^{r-j}\sum_{l=0}^{i+j-1} (\ast)=\sum_{l=0}^{r-1}\sum_{i=l-j+1}^{r-j}(\ast),$$ where $(\ast)$ is the term you're summing. (See also: https://math.stackexchange.com/questions/1876828/how-to-change-the-order-of-summation) Is this what you wanted? If so, I'll post an answer. – B. Núñez Jun 04 '20 at 17:22
  • @B.Núñez I appreciate your answer – Oily Jun 04 '20 at 18:18

1 Answers1

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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.

Markus Scheuer
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