<s>[INST] How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$
 [/INST] Let $H_n=\sum_{k=1}^n \frac1k$ denote the $n$-th harmonic number.
It is well known that $H_n \sim \ln n + \gamma$, where $\gamma$ is Euler-Mascheroni constant. (To be more precise, $\lim\limits_{n\to\infty} (H_n-\ln n) = \gamma$, which is what we actually need below.)
You are interested in 
$$\sum_{k=1}^n \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac3{3k}\right) = H_{3n}-H_n \sim \ln{3n}-\ln n = \ln 3.$$
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