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: $$ \sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2 \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}} $$

What do you get if you differentiate $$\sum_{n=0}^\infty \frac{x^{3n+2}}{(3n+1)(3n+2)}$$ twice?

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

$$ x+\frac{1}{2} = \frac{\sqrt{3}}{2} tan \theta$$

$$ dx = \frac{\sqrt{3}}{2} sec^{2} \theta$$

$$ \begin{eqnarray} \int \frac{dx}{1+x+x^2} = \int \frac{ \frac{\sqrt{3}}{2} sec^{2} \theta}{\frac{3}{4} sec^{2} \theta} {\mathrm{d} \theta} &=& \frac{2}{\sqrt{3}} \theta \\ &=& \frac{2}{\sqrt{3}} tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) \end{eqnarray} $$

$$ \Rightarrow \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \frac{2\sqrt{3}}{3} \left( tan^{-1} ( \frac{3}{\sqrt{3}} ) - tan^{-1} ( \frac{1}{\sqrt{3}} ) \right)$$

$$ = \frac{2\sqrt{3}}{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}}$$