Evaluate $\sum_{n=0}^\infty (-1)^n \ln\frac{2+2n}{1+2n}$ 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.
 A: We have the series expansion of the digamma function:
$$\psi(z)=-\gamma+\sum_{n=0}^\infty\frac1{n+1}-\frac1{n+z}$$
Integrating both sides gives
$$\ln\Gamma(z)=(1-z)\gamma+\sum_{n=0}^\infty\frac{z-1}{n+1}+\ln\left[\frac{n+1}{n+z}\right]$$
Hence we have
$$\ln\Gamma(1)+\ln\Gamma(1/4)-\ln\Gamma(1/2)-\ln\Gamma(3/4)=\sum_{n=0}^\infty\ln\left[\frac{(n+1/2)(n+3/4)}{(n+1)(n+1/4)}\right]$$

$$\ln\left[\frac{\Gamma(1/4)}{\Gamma(3/4)}\right]-\frac12\ln(\pi)=\sum_{n=0}^\infty\ln\left[\frac{(4n+2)(4n+3)}{(4n+4)(4n+1)}\right]$$

A: 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 $\Gamma\left(\frac{3}{4}\right)=\frac{\sqrt{2}\pi }{\Gamma\left(\frac{1}{4}\right)}$.
A: $$ 
\begin{align} 
\color{red}{S}&=\sum_{n=0}^{\infty}(-1)^{n}\,\log\left[\frac{2n+2}{2n+1}\right] \\[2mm] 
&=\log\prod_{n=0}^{\infty}\left[\frac{\left(4n+2\right)\left(4n+3\right)}{\left(4n+1\right)\left(4n+4\right)}\right] \\[2mm] 
&=\log\prod_{n=0}^{\infty}\left[\frac{\left(1+\frac{2/4}{\Large{n}}\right)\left(1+\frac{3/4}{\Large{n}}\right)}{\left(1+\frac{1/4}{\Large{n}}\right)\left(1+\frac{4/4}{\Large{n}}\right)}\right] \\[2mm] 
&=\log\prod_{n=0}^{\infty}\left[\frac{\left(1+\frac{2/4}{\Large{n}}\right)}{{\left(1+\frac{1}{\Large{n}}\right)}^{\small{2/4}}}\frac{\left(1+\frac{3/4}{\Large{n}}\right)}{{\left(1+\frac{1}{\Large{n}}\right)}^{\small{3/4}}}\frac{{\left(1+\frac{1}{\Large{n}}\right)}^{\small{1/4}}}{\left(1+\frac{1/4}{\Large{n}}\right)}\right] \\[2mm] 
&=\color{red}{\log\left[\frac{\Gamma\left(1/4\right)}{\Gamma\left(2/4\right)\,\Gamma\left(3/4\right)}\right]} 
\end{align} 
$$ 
