Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$ I would like to show that:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$
We have:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$
I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...

Regularized the series:
$$ \begin{eqnarray}
   \sum_{n=0}^m \frac{1}{(3n+1)(3n+2)} &=& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) = \sum_{n=0}^m  \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
    &=& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x =  
         \int_0^1  \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x
\end{eqnarray}
$$
Now we can take the limit by dominating convergence theorem:
$$
  \sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2   \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}}
$$
