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Let $a \notin \mathbb{Z}$ and $a \neq \frac{1}{2}$. Prove that

$$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$$

Attempt

Using the fact that

\begin{align*} \frac{1}{\Gamma\left ( a+x \right ) \Gamma \left ( \beta - x \right )} &= \frac{1}{\left ( a+x-1 \right )! \left ( \beta-x-1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \frac{\left ( a + \beta-2 \right )!}{\left ( a + x -1 \right )! \left ( \beta - x -1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \binom{a + \beta - 2}{a + x -1} \end{align*}

the question really boils down to the sum

$$\mathcal{S} = \sum_{n=0}^{\infty} \binom{2a-2}{a+n-1}$$

To this end,

\begin{align*} \sum_{n=0}^{\infty} \binom{2a-1}{a+n-1} &=\frac{1}{2\pi i} \sum_{n=0}^{\infty} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a+n}}\, \mathrm{d}z \\ &= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1 + z \right )^{2a-1}}{z^a} \sum_{n=0}^{\infty} \frac{1}{z^n} \, \mathrm{d}z \\ &= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a-1} \left ( z-1 \right )} \, \mathrm{d}z \end{align*}

using the handy identity $\displaystyle \binom{n}{k} = \frac{1}{2\pi i } \oint \limits_{\gamma} \frac{\left ( 1+z \right )^n}{z^{k+1}} \, \mathrm{d}z$. I think I'm on the right track, but I'm having a difficult time evaluating the last contour integral. Any help?

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    $\begingroup$ Since you assume that $a$ is not an integer, you have problems of having $z^a$ around $z=0$ in the integrand. It is not an analytic function around the origin. $\endgroup$
    – Gary
    May 9, 2022 at 11:28
  • $\begingroup$ Yeah , I can see that! But unfortunately this is how the question is formulated !!! Any other ideas ? $\endgroup$
    – Tolaso
    May 9, 2022 at 11:31
  • $\begingroup$ What does $\binom{2a-2}{a+n-1}$ mean if $a+n-1$ is not a (non-negative) integer? $\endgroup$
    – Martin R
    May 9, 2022 at 13:30
  • $\begingroup$ @MartinR It seems if $2a - 2$ is a negative integer (e.g. $a = -3/2$), then $\binom{2a-2}{a+n-1}$ is not well-defined, according to the definition $\binom{z}{w} = \frac{\Gamma(z+1)}{\Gamma(w+1)\Gamma(z-w+1)}$. $\endgroup$
    – River Li
    May 10, 2022 at 3:33
  • $\begingroup$ @MartinR I think a condition should be added: $2a - 1$ is not a negative integer. $\endgroup$
    – River Li
    May 10, 2022 at 3:59

3 Answers 3

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If $a\notin\mathbb{Z}$ then the sum converges if and only if $\color{red}{a>1/2}$, which is obtained from $$\lim_{x\to+\infty}\frac{x^\lambda\Gamma(x)}{\Gamma(\lambda+x)}=1\implies\lim_{n\to\infty}\frac{(-1)^n\color{red}{n^{2a-1}}}{\Gamma(a+n)\Gamma(a-n)}=\frac{\sin a\pi}{\pi}$$ using the reflection formula for $\Gamma(a-n)$.

In this case, a possible solution is to apply the Poisson summation formula $$\sum_{n\in\mathbb{Z}}f(n)=\sum_{n\in\mathbb{Z}}\hat{f}(n),\qquad\hat{f}(y):=\int_{-\infty}^\infty f(x)e^{-2i\pi xy}\,dx$$ to $f(x)=\cos^{2(a-1)}\pi x$ for $|x|<1/2$ (and $f(x)=0$ elsewhere); then we get $$\hat{f}(y)=\frac{2^{2-2a}\Gamma(2a-1)}{\Gamma(a+y)\Gamma(a-y)}$$ as a "reciprocal beta" integral (see DLMF or e.g. this question), and the result follows.

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  • $\begingroup$ I don’t see how $\frac{1}{\Gamma^2(a)}$ appears. Nice approach using the Poisson summation formula!! $\endgroup$
    – Tolaso
    May 10, 2022 at 11:28
  • $\begingroup$ @Tolaso: $\sum\limits_{\color{red}{n\in\mathbb{Z}}}\frac1{\Gamma(a+n)\Gamma(a-n)}=2\sum\limits_{n=0}^\infty\frac1{\Gamma(a+n)\Gamma(a-n)}-\frac1{\Gamma^2(a)}$. $\endgroup$
    – metamorphy
    May 10, 2022 at 15:16
  • $\begingroup$ Yeah, that is right. $\endgroup$
    – Tolaso
    May 11, 2022 at 17:23
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To supplement Claude Leibovici's answer, I will use Euler's integral representation of the Gaussian hypergeometric function to show that $$_2F_1(1,1-a;a;-1)=\frac{1}{2} \left(\frac{2^{2a-2} \, \Gamma^{2}(a)}{\Gamma(2a-1)} +1 \right) $$ for at least $a >1$. This is where the series $\sum_{n=0}^{\infty} \frac{2}{\Gamma ( a + n) \Gamma ( a - n )} $ converges absolutely.

$ \begin{align} _2F_1(1,1-a;a;-1) &= \, _2F_1(1-a,1;a;-1) \\ &= \frac{1}{B(1,a-1)} \int_{0}^{1} (1-x)^{a-2} (1+x)^{a-1} \, \mathrm dx \\ &= (a-1) \int_{0}^{1} (1-x)^{a-2}(1+x)^{a-2}(1+x) \, \mathrm dx\\ &= (a-1) \left( \int_{0}^{1} (1-x^{2})^{a-2} \, \mathrm dx +\int_{0}^{1} x (1-x^{2})^{a-2} \, \mathrm dx \right) \\ &= (a-1) \left( \frac{1}{2} \int_{0}^{1} (1-u)^{a-2} u^{-1/2} \, \mathrm du + \frac{1}{2}\int_{0}^{1} v^{a-2} \, \mathrm dv \right) \\ &= (a-1) \left( \frac{1}{2} \, B \left(\frac{1}{2},a-1\right) + \frac{1}{2(a-1)} \right) \\ &= \frac{a-1}{2} \left(\frac{ \sqrt{\pi} \, \Gamma(a-1)}{\Gamma \left(a- \frac{1}{2}\right)}+\frac{1}{a-1} \right) \\ &\overset{(1)}= \frac{a-1}{2} \left(\frac{\sqrt{\pi} \, \Gamma(a) 2^{2(a-1/2)-1} \Gamma \left(a-\frac{1}{2}+\frac{1}{2} \right)}{(a-1) \Gamma \left( 2\left(a-\frac{1}{2}\right)\right) \sqrt{\pi}}+\frac{1}{a-1} \right) \\ &= \frac{1}{2} \left(\frac{2^{2a-2} \Gamma^{2}(a)}{\Gamma(2a-1)} +1 \right). \end{align}$


$(1)$ Legendre duplication formula

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If you enjoy special functions $$f(x)=\sum_{n=0}^{\infty} \frac{2\,x^n}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2 }{\Gamma (a)^2}\,\, _2F_1(1,1-a;a;-x)$$ $$\, _2F_1(1,1-a;a;-1)=\frac{1}{2} \left(1+\sqrt{\pi }\frac{ \Gamma (a)}{\Gamma \left(a-\frac{1}{2}\right)}\right)$$ $$f(1)=\frac{\sqrt{\pi }}{\Gamma \left(a-\frac{1}{2}\right) \Gamma (a)}+\frac{1}{\Gamma (a)^2}$$

Now, work the first term to get the desired result.

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    $\begingroup$ How is that result for $_2F_1(1,1-a;a;-1)$ derived $\endgroup$
    – FShrike
    May 9, 2022 at 12:11

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