<s>[INST] 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$...
 [/INST] An important trick here is that sigma and integral signs can be changed around. 
$$\int \sum^b_{n=a} f\left(n,x\right)\, dx = \sum^b_{n=a} \int f\left(n,x\right) \,dx$$
And this is because
$$\int \sum^b_{n=a} f(n,x)\, dx$$
$$\int f\left(a,x\right) + f((a+1),x) + f((a+2),x) + \dots +f\left((b-1),x\right) + f(b,x) $$
$$ = \int f(a,x)\,dx + \int f((a+1),x) \,dx + \dots + \int f((b-1),x)\, dx + \int f(b,x)\, dx$$
Therefore 
$$\begin{align*}
   \sum_{n=0}^m \frac{1}{(3n+1)(3n+2)} 
    =& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) \\
    =& \sum_{n=0}^m  \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
    =& \int_0^1 \sum_{n=0}^m \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
    =& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x =  
         \int_0^1  \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x
\end{align*}
$$
Also because 
$$ \sum_{n=0}^m \left( x^{3n} - x^{3n+1} \right) = \frac{(1-x^{3(m+1)})(1-x)}{1-x^3} $$
Now let us see how the final integral 
$$\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} $$
is evaluated.
$$ x^2+x+1 = \left(x+\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2$$
therefore if you skip two steps of substitution and do it once 
$$ x+\frac{1}{2} = \frac{\sqrt{3}}{2} tan \theta$$
$$ dx = \frac{\sqrt{3}}{2} sec^{2} \theta$$
$$
\begin{eqnarray}
\int \frac{dx}{1+x+x^2} =  \int \frac{ \frac{\sqrt{3}}{2} sec^{2} \theta}{\frac{3}{4} sec^{2} \theta} {\mathrm{d} \theta}
   &=& \frac{2}{\sqrt{3}} \theta \\
   &=& \frac{2}{\sqrt{3}} tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) 
\end{eqnarray}
$$
$$ \Rightarrow \int_0^1  \frac{\mathrm{d} x}{1+x+x^2} = \frac{2\sqrt{3}}{3} \left( tan^{-1} ( \frac{3}{\sqrt{3}} ) - tan^{-1} ( \frac{1}{\sqrt{3}} ) \right)$$
$$ = \frac{2\sqrt{3}}{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}}$$
</s>