How to prove that $\sum_{n=0}^{\infty} (-1)^n \ln \frac{3n+2}{3n+1}=\frac{1}{2} \ln 3$ The sum $$\sum_{n=0}^{\infty} (-1)^n \ln \frac{3n+2}{3n+1}=\frac{1}{2} \ln 3$$ has been encountered in the post below:
How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$
I would like to know as to how this sum can be proved.

We can obtain something analogous to the Wallis product using$$\prod_{n\ge1}(1-z^2/n^2)=\frac{\sin\pi z}{\pi z}\implies\prod_{n\ge1}(1-z^2/(2n+1)^2)=\frac{\frac{\sin\pi z}{\pi z}}{\frac{\sin\pi z/2}{\pi z/2}}=\cos\frac{\pi z}{2},$$so$$\prod_{n\ge0}\frac{(6n+2)(6n+4)}{(6n+1)(6n+5)}=\prod_{n\ge0}\frac{1-1/(6n+3)^2}{1-4/(6n+3)^2}=\frac{\cos\frac{\pi}{6}}{\cos\frac{\pi}{3}}=\sqrt{3}.$$
