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.
 A: 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$$\prod_{n\ge0}\frac{(6n+2)(6n+4)}{(6n+1)(6n+5)}=\prod_{n\ge0}\frac{1-1/(6n+3)^2}{1-4/(6n+3)^2}=\frac{\cos\frac{\pi}{6}}{\cos\frac{\pi}{3}}=\sqrt{3}.$$
A: To evaluate the series we appeal to the digamma function, its relationship with the Gamma function, and Euler's reflection formula.  Proceeding, we write
$$\begin{align}
\sum_{n=0}^\infty (-1)^n\log\left(\frac{3n+2}{3n+1}\right)&=\int_0^1 \sum_{n=0}^\infty (-1)^n \frac1{s+3n+1}\,ds\\\\
&=\int_0^1 \sum_{n=0}^\infty\left(\frac1{6n+s+1}-\frac1{6n+s+4}\right)\,ds\\\\
&=\frac16\int_0^1\left(\psi((s+4)/6)-\psi((s+1)/6)\right)\,ds\\\\
&=\log\left(\frac{\Gamma(5/6)\Gamma(1/6)}{\Gamma(2/3)\Gamma(1/3)}\right)\\\\
&=\log\left(\frac{\sin(2\pi/3)}{\sin(5\pi/6)}\right)\\\\
&=\log(\sqrt 3)
\end{align}$$
as expected!
A: 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.
