Calculate the sum $\sum_{n=0}^{\infty}\frac{1}{3n+2}-\frac{1}{3n+3}$.
There was this hint to use Mittag-Leffler decomposition of $\frac{\pi}{a}\cot{\frac{\pi b}{a}}=\sum_{n=0}^{\infty}(\frac{1}{na+b}-\frac{1}{na+{a-b}})$. So I found this useful to calulate another sum, namely: $\sum_{n=0}^{\infty}\frac{1}{3n+1}-\frac{1}{3n+2}=\frac{\pi\sqrt{3}}{9}$. But I can't see how it can be used to the question related sum. Any other approach is welcome.
We can rewrite the summation as an integral:
$$\sum_{n=0}^\infty \frac{1}{3n+2}-\frac{1}{3n+3} = \int_0^1\sum_{n=0}^{\infty}x^{3n+1}-x^{3n+2}\:dx$$
$$= \int_0^1\frac{x}{1-x^3}-\frac{x^2}{1-x^3}\:dx = \int_0^1 \frac{x}{1+x+x^2}\:dx$$
Then to calculate the integral, add and subtract $\frac{1}{2}$ in the numerator
$$\int_0^1 \frac{x+\frac{1}{2}-\frac{1}{2}}{1+x+x^2}\:dx = \frac{1}{2}\int_0^1 \frac{1+2x}{1+x+x^2}\:dx - \frac{1}{2}\int_0^1 \frac{1}{\frac{3}{4}+\left(\frac{1}{2}+x\right)^2}\:dx$$
$$= \frac{1}{2}\log(1+x+x^2)-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{1+2x}{\sqrt{3}}\right)\Biggr|_0^1 = \log\sqrt{3} - \frac{\pi}{6\sqrt{3}}$$
$$S= \sum_{n=0}^{\infty}\int (x^{3n+1}-x^{3n+2})dx= \int_{0}^{1} \sum_{n=0}^{\infty} [x(x^3)^n-x^2(x^3)^{n}] dx=\int_{0}^{1} \frac{x}{1+x+x^2} dx=\frac{\ln 3}{2}-\frac{\sqrt{3}\pi}{18}.$$ Here we have used integral representation of $\frac{1}{t}=\int_{0}^{1}x^{t-1}dx$ and sum if iGP.
