Find the closed form of the gamma function related series I see a possible way of computing  the series by using integrals, but I wonder if it possible
to avoid the use of them, to get a neat evaluation by only using series.
Compute
$$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(\frac{\displaystyle \Gamma\left(\frac{n+2}{2}\right)\displaystyle\Gamma\left(\frac{n}{2}\right)}{\left(\displaystyle\Gamma\left(\frac{n+1}{2}\right)\right)^2}\right)$$
And here is a supplementary question
$$\sum_{n=1}^{\infty} (-1)^{n+1}\psi^{(0)}(\alpha n)\log\left(\frac{\displaystyle \Gamma\left(\frac{n+2}{2}\right)\displaystyle\Gamma\left(\frac{n}{2}\right)}{\left(\displaystyle\Gamma\left(\frac{n+1}{2}\right)\right)^2}\right),\space \alpha >0$$
 A: Decomposing the sum into two parts, with $n=2k$ and $n=2k-1$ and recombining them, we can rewrite it as
$$S=\sum_{k=1}^{\infty}\ln\frac{\Gamma^3\left(k+\frac12\right)\Gamma\left(k-\frac12\right)}{\Gamma^3\left(k\right)\Gamma\left(k+1\right)} \tag{1}$$
Next, using the recursion relation for the Barnes $G$-function, $G(z+1)=\Gamma(z)G(z)$, a finite analog of the sum (1) can be computed:
$$S_N=\sum_{k=1}^{N-1}\ln\frac{\Gamma^3\left(k+\frac12\right)\Gamma\left(k-\frac12\right)}{\Gamma^3\left(k\right)\Gamma\left(k+1\right)}=\ln\frac{G^3\left(N+\frac12\right)G\left(N-\frac12\right)G^3\left(1\right)G\left(2\right)}{G^3\left(\frac32\right)G\left(\frac12\right)G^3\left(N\right)G\left(N+1\right)}.$$
Now to compute the original sum $S=S_{\infty}$, it suffices to use the known asymptotics of $G(z)$ for large argument. This gives
$$\ln\frac{G^3\left(N+\frac12\right)G\left(N-\frac12\right)}{G^3\left(N\right)G\left(N+1\right)}=O\left(\frac1N\right),$$
and in turn implies
$$S=-\ln\left(G^3\left(\frac32\right)G\left(\frac12\right)\right)=-\ln\left(\pi^{\frac32}G^4\left(\frac12\right)\right)=-6\zeta'\left(-1\right)-\frac{\ln 2}{6}-\frac{\ln \pi}{2}.$$
The final step uses the known evaluation of $G\left(\frac12\right)$, see the formula (19) here.
A: $$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(\frac{\displaystyle \Gamma\left(\frac{n+2}{2}\right)\displaystyle\Gamma\left(\frac{n}{2}\right)}{\left(\displaystyle\Gamma\left(\frac{n+1}{2}\right)\right)^2}\right)=\log\left(\frac{A^6}{2^{1/6}\sqrt{\pi e}}\right)$$
