<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] Use the fact that $ \displaystyle \frac{1}{m+1} = \int^1_0 x^m dx$ to get $$ \sum_{n=1}^{\infty} \int^1_0 \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$
$$ = \int^1_0 \sum_{n=1}^{\infty} \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$
$$= \int^1_0 \frac{ x+1-2x^2}{1-x^3} dx= \int^1_0 \frac{2x+1}{x^2+x+1} dx  = \log (x^2+x+1) \biggr|^1_0=\log 3.$$
</s>