Proving $\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a)$ 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
 A: 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:
$\Re \int\limits_0^1\frac{\Gamma(a+1)}{(1- it)^{a+1}}dt=\Re \int\limits_{1-i}^1\frac{\Gamma(a+1)}{ix^{a+1}}dx=\frac {\Gamma(a+1)}{a}\sqrt{2}^{-a}\sin(\frac{\pi}{4}a)$
A: 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
A: 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
$$\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$$
$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$
$$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$
