Evaluate $\sum_{n=0}^\infty (-1)^n \ln\frac{2+2n}{1+2n}$

Question

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.

Answer 1

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

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

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