TITLE: Evaluate $\sum_{n=0}^\infty (-1)^n \ln\frac{2+2n}{1+2n}$
QUESTION [6 upvotes]: I'm trying to evaluate:
$$\sum_{n=0}^\infty (-1)^n \ln\frac{2+2n}{1+2n}$$
I'm not too sure where to start. I've tried writing it as a telescoping sum, but that doesn't work. I'm thinking it's related to the Wallis Product or some infinite product of Gamma's somehow, but I can't figure anything out.
I've rewritten it as:
$$\sum_{n=0}^\infty \ln\frac{(2+4n)(3+4n)}{(1+4n)(4+4n)}=\ln\prod_{n=0}^\infty\frac{(2+4n)(3+4n)}{(1+4n)(4+4n)}$$
but to no avail.

REPLY [3 votes]: Just want to point out that, a general version of this exist (Almodavar, Hall et.al https://cs.uwaterloo.ca/journals/JIS/VOL19/Moll/moll3.pdf). 
\begin{eqnarray*}
\prod_{n=0}^{\infty}{\left(\frac{2\alpha n+\beta}{2\alpha n+\gamma}\right)^{(-1)^{n}}} &=& 2^{\frac{\gamma-\beta}{2\alpha}} \frac{\Gamma^{2}\left(\frac{\gamma}{4\alpha}\right)}{\Gamma^{2}\left(\frac{\beta}{4\alpha}\right)}   \frac{\Gamma\left(\frac{\beta}{2\alpha}\right)}{\Gamma\left(\frac{\gamma}{2\alpha}\right)}
\end{eqnarray*}
Taking $\ln$ then will have,
\begin{eqnarray*}
\sum_{n=0}^{\infty}(-1)^{n} {\ln \left(\frac{2\alpha n+\beta}{2\alpha n+\gamma}\right)} &=& \sum_{n=0}^{\infty}{\ln \left(\frac{2\alpha n+\beta}{2\alpha n+\gamma}\right)^{(-1)^{n}}} \\
&=& \ln \prod_{n=0}^{\infty}{\left(\frac{2\alpha n+\beta}{2\alpha n+\gamma}\right)^{(-1)^{n}}} \\
&=& \ln \left(2^{\frac{\gamma-\beta}{2\alpha}} \frac{\Gamma^{2}\left(\frac{\gamma}{4\alpha}\right)}{\Gamma^{2}\left(\frac{\beta}{4\alpha}\right)}   \frac{\Gamma\left(\frac{\beta}{2\alpha}\right)}{\Gamma\left(\frac{\gamma}{2\alpha}\right)}\right)
\end{eqnarray*}
The question of interest here is a case of $\alpha=1,\beta=2,$ and $\gamma=1$, which will then become, 
\begin{eqnarray*}
\sum_{n=0}^{\infty}(-1)^{n} {\ln \left(\frac{2 n+2}{2 n+1}\right)} &=& \ln \left(2^{-\frac{1}{2}} \frac{\Gamma^{2}\left(\frac{1}{4}\right)}{\Gamma^{2}\left(\frac{1}{2}\right)}   \frac{\Gamma\left(1\right)}{\Gamma\left(\frac{1}{2}\right)}\right) \\
&=& \ln \left(  \frac{1}{\sqrt{2 \pi}}\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{\pi} \right) \\
&\approx& 0.512377 
\end{eqnarray*}
This is the same result as what is proved neatly by SimplyBeautifulArt. Recall that $\Gamma\left(\frac{3}{4}\right)=\frac{\sqrt{2}\pi }{\Gamma\left(\frac{1}{4}\right)}$.