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Mathematica is able to calculate the following sum:

\begin{align} &\sum_{k,l=0}^{n,m}\frac{(-1)^{n+m-l-k}(2m)!(2n+1)!(2 )^{2k+2l}}{(2k+1)!(2l)!(n-k)!(m-l)!}(k+l)! \nonumber\\ =&\frac{(-1)^{m+n}2^{2m+2n+1}}{\pi(2n+1-2m)}\Gamma(m+\frac 1 2)\Gamma(n+1+\frac 1 2) \nonumber\\ =&\frac{(-2) ^{m+n}(2m-1)!!(2n+1)!!}{(2n+1-2m)} \end{align}

(The last step is just a way of re-writing it in terms of double factorial.) I tried proving it (without putting in too much effort, however) but was not succesful. Do you have hints how this should be approached (or even full solutions)? I tried complete induction but ended up at a point where I would have had to prove a slightly less difficult summation formula.

Gesbesgue
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2 Answers2

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We consider the identity in the form \begin{align*} &\sum_{k,l=0}^{n,m}\frac{(-1)^{n+m-l-k}(2m)!(2n+1)!2^{2k+2l}}{(2k+1)!(2l)!(n-k)!(m-l)!}(k+l)!=\frac{(-2) ^{m+n}(2m-1)!!(2n+1)!!}{2n+1-2m} \end{align*}

We divide both sides by $(-2)^{m+n}(2m-1)!!(2n+1)!!$ and obtain \begin{align*} \frac{(2m)!(2n+1)!}{2^{m+n}(2m-1)!!(2n+1)!!}\sum_{k,l=0}^{n,m}\frac{(-4)^{k+l}(k+l)!}{(2k+1)!(2l)!(n-k)!(m-l)!} =\color{blue}{\frac{1}{2n+1-2m}} \end{align*} We start with the left-hand side and we obtain \begin{align*} &\frac{(2m)!(2n+1)!}{2^{m+n}(2m-1)!!(2n+1)!!}\sum_{k,l=0}^{n,m}\frac{(-4)^{k+l}(k+l)!}{(2k+1)!(2l)!(n-k)!(m-l)!}\\ &\qquad=\frac{(2m)!!(2n)!!}{2^{m+n}}\sum_{l=0}^m\frac{(-4)^l}{(2l)!(m-l)!}\sum_{k=0}^n\frac{(-4)^k(k+l)!}{(2k+1)!(n-k)!}\tag{1}\\ &\qquad=m!n!\sum_{l=0}^m\frac{(-4)^l}{(m-l)!l!l!}\binom{2l}{l}^{-1} \sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{2k}{k}^{-1}\frac{(k+l)!}{(n-k)!k!k!}\tag{2}\\ &\qquad\color{blue}{=\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1} \sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}} \end{align*}

The equality of the last line with $\color{blue}{\frac{1}{2n+1-2m}}$ is given in this MSE post.

Comment:

  • In (1) we use the identity $n!=n!!(n-1)!!$

  • In (2) we use $(2n)!!=2^nn!$

Markus Scheuer
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  • Thanks a lot - this really shows the benefits this page has. I also learned new methods from the different solutions in your linked post. – Gesbesgue Aug 17 '17 at 09:30
  • @Gesbesgue: You're welcome! :-) You might also be interested to take a somewhat closer look at the comments of my answer. It explicitly states two somewhat advanced representations of binomial coefficients $\binom{n}{k}$ as integral and as residue. – Markus Scheuer Aug 17 '17 at 09:34
  • Thanks again @Markus, I studied them today :) – Gesbesgue Sep 20 '17 at 20:19
  • @Gesbesgue: You're welcome! :-) – Markus Scheuer Sep 20 '17 at 20:49
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I do not think simple induction will work. I have reduced it to the identity $$ 1 = \sum_{k=0}^n \sum_{j=0}^m \frac{(-4)^{k+j}(k+j)!(n)!(m)!(2n+1-2m)} {(2k+1)!(2j)!(n-k)!(m-j)!} $$ for all $n\ge 0, m\ge 0$. The Wilf-Zeilberger method could prove this but it is not satisfying to me.

Somos
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