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.

Not a solution but a starting point. Use that $\ln(x)$ has the property that $\ln(ab) = \ln(a) + \ln(b)$. So we can rewrite the partial sum as $$\sum_{n=0}^N (-1)^n\ln\Big(\frac{3n+2}{3n+1}\Big) = \ln\Big(\prod_{n=0}^N \Big(\frac{3n+2}{3n+1}\Big)^{(-1)^n}\Big)$$
Since $x\to \ln(x)$ is a continuous function, then if we can show that
$$ \prod_{n=0}^N \Big(\frac{3n+2}{3n+1}\Big)^{(-1)^n} \to \sqrt{3}$$
as $N\to\infty$, then we will have the result. It looks J.G. answer above will be helpful for this part.
