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Mathematica gives

$$\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a),\quad 0<a<1$$

All I did is reindexing then using the series property $\sum_{n=1}^\infty (-1)^n f(2n)=\Re \sum_{n=1}^\infty i^n f(n)$ ;

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

and I dont know how to continue, any idea?

Thanks

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    $\begingroup$ It is amazing to me to see this problem today. We worked the partial sums many years ago. $\endgroup$ – Claude Leibovici Jun 30 '20 at 9:21
  • $\begingroup$ Thanks @Claude Leibovici .. Glad you like it. $\endgroup$ – Ali Shadhar Jun 30 '20 at 11:51
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A solution in large steps by Cornel Ioan Valean

In the following, I'll focus on the last series. Let's prove that

$$\sum_{n=1}^{\infty} x^n \frac{\Gamma(n+a-1)}{\Gamma(n)}=\frac{x}{(1-x)^a}\Gamma(a).$$

Two key steps are necessary:

$1)$. Note and use that

$$\frac{1}{\Gamma(1-a)}\int_0^1 t^{-a} (1-t)^{n+a-2}\textrm{d}t=\frac{\Gamma(n+a-1)}{\Gamma(n)}.$$

$2)$. (after summing) Employ the following integral representation with a hypergeometric structure (in fact, it may be viewed as a particular case of an integral expressed in terms of a hypergeometric function)

$$\int_0^1 \frac{x^{a-1}}{(1-x)^a (1+b x)}\textrm{d}x=\frac{\pi}{\sin(\pi a)}\frac{1}{(1+b)^a}.$$

One useful way to perform the evaluation of the last integral is by using the variable change $x/(1-x)=y$, followed by the variable change $(1+b)y=z$ in order to get precisely a special case of the Beta function.

End of story

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    $\begingroup$ Fully understood ... thanks to you and Cornel for such amazing work. Big(+1). $\endgroup$ – Ali Shadhar Jun 30 '20 at 11:05
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Let's transform it into a factorial form

$\sum_\limits{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=\sum_\limits{n=0}^\infty\frac{(-1)^n(2n+a)!a!}{(2n+1)!a!}=\sum_\limits{n=0}^\infty\frac{(i)^{2n}\binom{2n+a}{2n}a!}{2n+1}$

After reindexing and using the mentioned series property we have:

$\sum_\limits{n=0}^\infty\frac{(i)^{2n}\binom{2n+a}{2n}a!}{2n+1}=\Re \sum_\limits{n=0}^\infty\frac{(i)^{n}\binom{n+a}{n}a!}{n+1}$

Apply the fact $\frac{1}{n+1}=\int\limits_0^1 t^n dt$ and the binomial identity: $\frac{1}{(1-z)^{a+1}}=\sum\limits_{n=0}^\infty \binom{n+a}{n}z^a$

We get:

$\Re \int\limits_0^1\frac{\Gamma(a+1)}{(1- it)^{a+1}}dt=\Re \int\limits_{1-i}^1\frac{\Gamma(a+1)}{ix^{a+1}}dx=\frac {\Gamma(a+1)}{a}\sqrt{2}^{-a}\sin(\frac{\pi}{4}a)$

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  • $\begingroup$ very nice and simple .. thanks (+1). $\endgroup$ – Ali Shadhar Jun 30 '20 at 14:14
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    $\begingroup$ It was my pleasure. $\endgroup$ – JV.Stalker Jun 30 '20 at 15:30
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This is not a solution, its just a bonus using the generalization derived by Cornel shown in the solution above.

He showed

$$\sum_{n=1}^{\infty} \frac{\Gamma(n+a-1)}{\Gamma(n)}x^{n-1} =\frac{\Gamma(a)}{(1-x)^a},\quad ,\quad 0<a<1$$

Set $a-1=-b$ we have

$$\quad \sum_{n=1}^\infty\frac{\Gamma(n-b)}{\Gamma(n)}x^{n-1}=\frac{\Gamma(1-b)}{(1-x)^{1-b}},\quad 0<b<1\quad \cdots\cdots(1)$$

$$i)\quad\sum_{n=1}^\infty\frac{\Gamma(n-b)}{n^2\ \Gamma(n)}=\frac{\Gamma(1-b)}{b}H_{b}$$

$$ii)\quad\sum_{n=1}^\infty\frac{\Gamma(n-b)}{n^3\ \Gamma(n)}=\frac{\Gamma(1-b)}{2b}\left(H_{b}^2+H_{b}^{(2)}\right)$$

$$iii)\quad\sum_{n=1}^\infty\frac{\Gamma(n-b)}{n^4\ \Gamma(n)}=\frac{\Gamma(1-b)}{6b}\left(H_{b}^3+3H_{b}H_{b}^{(2)}+2H_{b}^{(3)}\right)$$

$$iv)\quad\sum_{n=1}^\infty\frac{H_n \ \Gamma(n-b)}{n\ \Gamma(n)}=-\frac{\Gamma(-b)}{b}$$

where $i),\ ii)$ and $iii)$ follow from multiplying both sides of $(1)$ by $-\ln x,\ \frac12\ln^2x$ and $-\frac16\ln^3x$ respectively then $\int_0^1$ and $iv)$ follows from multiplying both sides of $(1)$ by $-\ln(1-x)$ then $\int_0^1.$

Also we used the following identities

$$\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$$

$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$

$$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$

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