How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$
 A: Let $s_N=\sum_{n=1}^N \frac1n$ be the $n$th partial sum of the harmonic series.
Then 
$$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}+\frac1{3n})-\sum_{n=1}^N\frac1n=s_{3N}-s_N.$$
It is well-known that $s_N=\ln N +\gamma+O\left(\frac1N\right)$, hence 
$$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\ln(3N)-\ln N+O\left(\frac1N\right)=\ln3+O\left(\frac1N\right).$$
A: 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.
A: 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 
$$\sum_{k=1}^n \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac3{3k}\right) = H_{3n}-H_n \sim \ln{3n}-\ln n = \ln 3.$$
A: Use the fact that $ \displaystyle \frac{1}{m+1} = \int^1_0 x^m dx$ to get $$ \sum_{n=1}^{\infty} \int^1_0 \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$
$$ = \int^1_0 \sum_{n=1}^{\infty} \left( x^{3n-2} + x^{3n-3} -2 x^{3n-1}\right) dx$$
$$= \int^1_0 \frac{ x+1-2x^2}{1-x^3} dx= \int^1_0 \frac{2x+1}{x^2+x+1} dx  = \log (x^2+x+1) \biggr|^1_0=\log 3.$$
