TITLE: How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$?
QUESTION [9 upvotes]: $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$

REPLY [6 votes]: Let's try one line proof using Riemann sums. So, our limit may be written as 
$$\lim_{n\to\infty}\sum_{k=1}^{n} \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac1k\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\frac1{1+\frac{k}{n}}=\int_0^2\frac{1}{1+x}=\ln3.$$
Q.E.D.