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I recently discovered the following infinite series

$\displaystyle \sum_{k=0}^{\infty} \binom{t}{k} (-1)^k \frac{(2k-3)!!}{(2k-2)!!} = \frac{2t \Gamma\left(t+\frac{1}{2}\right)}{\sqrt{\pi}{\Gamma(t+1)}} $

where $\displaystyle \binom{t}{k} = \frac{(t)_{k}}{k!} $ is the generalized binomial coefficient ($t$ is a real number). Any hint or guidance about the proof will be very appreciated.

It seems that the right hand side is related to the reciprocal of the complete beta function:

Since $\displaystyle B\left(t,\frac{1}{2}\right) = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma(t)}{\Gamma\left(t+\frac{1}{2}\right)} = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma(t+1)}{t\Gamma\left(t+\frac{1}{2}\right)}$

$\displaystyle \Longrightarrow \frac{1}{B\left(t,\frac{1}{2}\right)} = \frac{t\Gamma\left(t+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(t+1)}= \frac{t\Gamma\left(t+\frac{1}{2}\right)}{\sqrt{\pi}\Gamma(t+1)}$

So this sum is kind of an expansion series for the reciprocal of the complete beta function:

$\displaystyle \frac{2}{B\left(t,\frac{1}{2}\right)} = \sum_{k=0}^{\infty} \binom{t}{k} (-1)^k \frac{(2k-3)!!}{(2k-2)!!} $

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  • $\begingroup$ What is the question? $\endgroup$ Commented Aug 1, 2021 at 18:29
  • $\begingroup$ If someone has a hint about the proof $\endgroup$ Commented Aug 1, 2021 at 18:31

1 Answer 1

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Your series can be re-written as a limiting case of the Gauss hypergeometric function: $$ 2\mathop {\lim }\limits_{c \to 0} \frac{1}{{\Gamma (c)}}{}_2F_1 \!\left( { - t, - \tfrac{1}{2};c;1} \right). $$ You need to express the terms in terms of gamma functions to see this. Then, using http://dlmf.nist.gov/15.4.E20, we have $$ 2\mathop {\lim }\limits_{c \to 0} \frac{1}{{\Gamma (c)}}{}_2F_1 \!\left( { - t, - \tfrac{1}{2};c;1} \right) = 2\mathop {\lim }\limits_{c \to 0} \frac{{\Gamma \!\left( {c + t + \frac{1}{2}} \right)}}{{\Gamma (c + t)\Gamma \!\left( {c + \frac{1}{2}} \right)}} = 2\frac{{\Gamma \!\left( {t + \frac{1}{2}} \right)}}{{\Gamma (t)\Gamma \!\left( {\frac{1}{2}} \right)}} = \frac{{2t\Gamma\! \left( {t + \frac{1}{2}} \right)}}{{\sqrt \pi }\Gamma (t + 1)}. $$

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