10
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How to prove this: $$\sum_{k=0}^\infty \binom{x}{x-k}\cdot\binom{x}{k-x} = 1$$ For all $x\in\mathbb R_{\ge0}$ and with $\binom{x}{r}=\frac{\Gamma(x+1)}{\Gamma(r+1)\cdot\Gamma(x-r+1)}$

It is obviously true for all $x\in \mathbb N_0$, because then, all the values for $k≠x$ become $0$ and the one for $k=x$ becomes $1$. But after a few random calculations with Wolframalpha, I think it should hold for all positive real $x$.

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  • $\begingroup$ What does $n\choose r$ mean, when $r$ is not an integer? $\endgroup$ – Gerry Myerson Jul 14 '14 at 9:05
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    $\begingroup$ @GerryMyerson Perhaps a replacement of factorials with Gamma function values? $\endgroup$ – alex.jordan Jul 14 '14 at 9:14
  • $\begingroup$ Does $\binom{x}{r}$ mean $\frac{\Gamma(x+1)}{\Gamma(r+1)\cdot\Gamma(x-r+1)}$? $\endgroup$ – alex.jordan Jul 14 '14 at 9:17
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    $\begingroup$ @alex.jordan, yes, with $\binom{x}{r}$ I ment $\frac{\Gamma(x+1)}{\Gamma(r+1)*\Gamma(x-r+1)}$. Sorry, about that I was unclear. $\endgroup$ – Redundant Aunt Jul 14 '14 at 9:33
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    $\begingroup$ With $\verb*Clear[k, x]; Sum[ Binomial[x, x - k] Binomial[x, k - x], {k, 0, Infinity}, Assumptions -> x >= 0]*$, Mathematica yields $$ \frac{4^x \binom{x}{-x} \Gamma (1-x) \Gamma \left(\frac{1}{2} (2 x+1)\right)}{\sqrt{\pi }} = 1$$ $\endgroup$ – Felix Marin Sep 17 '14 at 4:35
5
+100
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For $x>0$, ${2x\choose k} = (-1)^k{k-2x-1\choose k}\sim \frac{(-1)^k}{k^{2x+1}\Gamma(-2x)}$, $k\to\infty$. Then the power series for $(1+z)^{2x}$ converges uniformly for $|z|\le 1$ by Weierstrass test; in particular, it can be integrated termwise. Therefore, $$ \sum_{k=0}^\infty \binom{x}{x-k}\cdot\binom{x}{k-x} = \sum_{k=0}^\infty \frac{\Gamma(x+1)^2}{\color{blue}{\Gamma(x-k+1)\Gamma(k-x+1)}\color{green}{\Gamma(2x-k+1)\Gamma(k+1)}} \\ = \frac{\Gamma(x+1)^2}{\color{green}{\Gamma(2x+1)}\color{blue}{\pi} }\sum_{k=0}^\infty\frac{\color{blue}{\sin \pi(k-x)}}{\color{blue}{k-x}}\color{green}{{2x \choose k}} = \frac{\color{red}{\Gamma(x+1)}^2}{\color{red}{\Gamma(2x+1)}\pi }\sum_{k=0}^\infty \mathrm{Re}\int_0^\pi e^{iy(k-x)}{2x \choose k}dy \\ = \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\mathrm{Re}\int_0^\pi e^{-i x y} \sum_{k=0}^\infty e^{iky}{2x \choose k}dy = \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\mathrm{Re}\int_0^\pi e^{-i x y} (1+e^{iy})^{2x}dy \\ = \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\mathrm{Re}\int_0^\pi \big(e^{-iy/2}(1+e^{iy})\big)^{2x}dy = \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\int_0^\pi \left(2\cos \frac{y}2\right)^{2x}dy\\ = \frac{2\Gamma(x+1)}{\Gamma(x+\frac12)\sqrt{\pi} }\int_0^{\pi/2} \left(\cos z\right)^{2x}dz = \frac{2\Gamma(x+1)}{\Gamma(x+\frac12)\sqrt{\pi} }\frac{\sqrt{\pi}\Gamma(x+\frac12)}{2\Gamma(x+1)} = 1. $$

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  • $\begingroup$ Nice! Would you mind justifying all the convergence and the interchanging of the integral and summation? $\endgroup$ – Hans Feb 3 '17 at 0:10
  • $\begingroup$ Yep. I just came back attempting to add the binomial expansion for justification only to find I was too late. :-) $\endgroup$ – Hans Feb 3 '17 at 9:05
  • $\begingroup$ $$ \sum_{n=0}^\infty\binom{s}{s-n}\cdot\binom{s}{n-s} = 1\quad\colon\,Re\{s\}\,\gt\,\color{red}{-\frac12} $$ Thanks. $\endgroup$ – Hazem Orabi Feb 8 '17 at 20:20

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