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Just started learning Gamma function, and we were asked to prove following equation for all positive integer $n$ and non-integer $m$.

$$0 = \sum^n_{i = 0}\frac{n-m-2i}{i!(n-i)!\Gamma (i+m+1) \Gamma (n-m-i+1)}$$

I tried when $n = 1$ and 2. I feel like it's related to an expansion of some binomial expression, but I can't figure out how to derive that expression. Maybe I'm in the wrong direction.

I searched online there's a generalized expression of binomial coefficient(not proved yet) Binomial formula for $(x+1)^{1/3}$ (related to Newton's binomial theorem)

$$\binom{n}{r} = \frac{\Gamma(n+1)}{\Gamma(r + 1)\Gamma(n-r + 1)}$$

Then the $$RHS = \frac{1}{n! \Gamma (n+1)} \sum^n_{i = 0}\binom{n}{m+i} \binom{n}{i} (n-m-2i)$$ or

$$ RHS = \frac{1}{n! \Gamma (n+1)} \sum^n_{i = 0}\binom{n}{m+i} \binom{n}{i} [n- (m+i) -i ]$$

I got stuck here and don't know what to do next.

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  • $\begingroup$ I suppose that you need to use the reflection formula for the gamma function. $\endgroup$ Commented Feb 29, 2020 at 6:20
  • $\begingroup$ You mean $\Gamma (1+p) \Gamma (1 - p) = \frac{\pi p}{sin(\pi p)}$? I tried, but I can't seem to cancel any of the term in the sum tho. $\endgroup$ Commented Mar 1, 2020 at 1:41
  • $\begingroup$ and I used $p\Gamma (p) = \Gamma (p+1)$ as well $\endgroup$ Commented Mar 1, 2020 at 2:22

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$\binom{\alpha}{\beta}:=\frac{\Gamma(\alpha+1)}{\Gamma(\beta+1)\Gamma(\alpha-\beta+1)}$ is well-defined (at least) for $\alpha\notin\mathbb{Z}_{<0}$ (assuming $1/\Gamma(\beta):=0$ for $\beta\in\mathbb{Z}_{\leq 0}$). Then $\binom{\alpha}{\beta}+\binom{\alpha}{\beta+1}=\binom{\alpha+1}{\beta+1}$ holds (like in the case of integers, and proven easily). Which allows to prove $$S(n,\alpha,\beta):=\sum_{k=0}^{n}\binom{n}{k}\binom{\alpha}{k+\beta}=\binom{n+\alpha}{n+\beta}$$ (extending Chu-Vandermonde identity) using induction on $n$. Now, since $$(n-m-i)\binom{n}{m+i}=n\binom{n-1}{m+i},\quad i\binom{n}{i}=n\binom{n-1}{i-1}\quad(i>0),$$ the given sum is equal to $\dfrac{S(n,n-1,m)-S(n-1,n,m+1)}{n!(n-1)!}=0.$

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  • $\begingroup$ hmm it looks about right. It's inspiring you distribute or separate the coefficient terms. I think that's what I should've considered. Thank you. $\endgroup$ Commented Mar 16, 2020 at 23:26

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