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$...
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\begin{align}&\color{#66f}{\large%
\sum_{n=0}^{\infty}\frac{1}{\pars{3n + 1}\pars{3n + 2}}}
=\frac{1}{9}\sum_{n=0}^{\infty}\frac{1}{\pars{n + 1/3}\pars{n + 2/3}}
=\frac{1}{3}\bracks{\Psi\pars{\frac{2}{3}} - \Psi\pars{\frac{1}{3}}}
\end{align}
where $\ds{\Psi}$ is the Digamma Function.
With Euler Reflection Formula:
\begin{align}&\color{#66f}{\large%
\sum_{n=0}^{\infty}\frac{1}{\pars{3n + 1}\pars{3n + 2}}}
=\frac{1}{3}\bracks{\pi\cot\pars{\pi\,\frac{1}{3}}}
=\color{#66f}{\large\frac{\pi}{3\root{3}}}
\end{align}
because $\ds{\cot\pars{\frac{\pi}{3}} = {1 \over \root{3}}}$.
A: 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:
$$
  \sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2   \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}}
$$
A: There is another way to solve the problem. Note
$$ (3n+1)(3n+2)=9\left[(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2\right] $$
and
and hence
\begin{eqnarray}
\sum_{n=0}^\infty\frac{1}{(3n+1)(3n+2)}&=&\frac19\sum_{n=0}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\
&=&\frac1{18}\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\
&=&\frac{1}{18}\frac{\pi\sinh2\pi b}{b(\cosh2\pi b-\cos2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{i}{6}}\\
&=&\frac{\pi}{3\sqrt3}
\end{eqnarray}
by using the result from this.
A: What do you get if you differentiate $$\sum_{n=0}^\infty \frac{x^{3n+2}}{(3n+1)(3n+2)}$$ twice?
A: 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 
$$ x+\frac{1}{2} = \frac{\sqrt{3}}{2} tan \theta$$
$$ dx = \frac{\sqrt{3}}{2} sec^{2} \theta$$
$$
\begin{eqnarray}
\int \frac{dx}{1+x+x^2} =  \int \frac{ \frac{\sqrt{3}}{2} sec^{2} \theta}{\frac{3}{4} sec^{2} \theta} {\mathrm{d} \theta}
   &=& \frac{2}{\sqrt{3}} \theta \\
   &=& \frac{2}{\sqrt{3}} tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) 
\end{eqnarray}
$$
$$ \Rightarrow \int_0^1  \frac{\mathrm{d} x}{1+x+x^2} = \frac{2\sqrt{3}}{3} \left( tan^{-1} ( \frac{3}{\sqrt{3}} ) - tan^{-1} ( \frac{1}{\sqrt{3}} ) \right)$$
$$ = \frac{2\sqrt{3}}{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}}$$
A: Since it seems to have gone unnoticed I would like to mention another common trick that can be used here, namely to study the function
$$f(z) = \frac{1}{(3z+1)(3z+2)} \pi\cot(\pi z)$$
which has the property that with $S$ being our sum,
$$2S = \sum_n \mathrm{Res}(f(z); z=n) = \sum_n \frac{1}{(3n+1)(3n+2)}.$$
We examine what happens when we integrate $f(z)$ along the rectangle
$$\Gamma =\pm (N+1/2) \pm i N$$
with $N$ a large positive integer.
There are no poles on this contour and seeing that $\frac{1}{(3z+1)(3z+2)}\in\Theta(1/N^2)$ on the contour, the integral $$\int_\Gamma f(z) dz$$ goes to zero as $N$ goes to infinity.
This implies that the sum of the residues at the poles of $f(z)$ is zero, giving (observe the two simple poles at $z=-1/3$ and $z=-2/3$)
$$2S + \frac{1}{3}\pi\cot(-\pi/3) - \frac{1}{3} \pi\cot(-2\pi/3) = 0$$
so that $$ S = \frac{\pi}{6} (\cot(-2\pi/3)-\cot(-\pi/3)) 
= \frac{\pi}{6} \frac{2}{\sqrt{3}} = \frac{\pi}{3\sqrt{3}}.$$
To convince yourself that the integral really does vanish consider the two lines $\Gamma_1$ which is $N+1/2\pm iN$ (right vertical) and $\Gamma_2$ which is $\pm N+1/2+iN$ (top horizontal).
We parameterize $\Gamma_1$ with $z=N+1/2+it$ so that
$$\left|\int_{\Gamma_1} f(z) dz \right|
= \left|\int_{-N}^N f(N+1/2+it) i dt\right|.$$
The fractional term attains its maximum at $t=0$ when we cross the real axis, giving an upper bound on the norm which is
$$\frac{1}{(3N+5/2)(3N+7/2)}.$$
For the norm of the trigonometric term we get
$$|\pi\cot(\pi (N+1/2) + \pi it)|
=\pi\left|\frac{e^{i\pi (N+1/2) - \pi t}+e^{-i\pi (N+1/2) + \pi t}}
{e^{i\pi (N+1/2) - \pi t}-e^{-i\pi (N+1/2) + \pi t}}\right|
\\ = \pi\left|\frac{i(-1)^N e^{- \pi t} - i(-1)^N e^{\pi t}}
{i(-1)^N e^{- \pi t} + i(-1)^N e^{\pi t}}\right|
= \pi|\tanh(\pi t)|.$$
Observe that with $t$ real $\tanh(\pi t)$ has no poles and its norm is bounded by one. Therefore the norm of the integral along $\Gamma_1$ is bounded by
$$2N \times \frac{1}{(3N+5/2)(3N+7/2)} \in \Theta(1/N)$$
and the integral vanishes as $N$ goes to infinity as claimed.
For $\Gamma_2$ we parameterize with $z = t + i N$ so that
$$\left|\int_{\Gamma_2} f(z) dz \right|
= \left|\int_{-(N+1/2)}^{N+1/2} f(t+iN) dt\right|.$$
The two factors of the fractional term are minimized when they cross the imaginary axis at $t=-1/3$ and $t=-2/3$, giving an upper bound on the norm which is
$$\frac{1}{3N\times 3N} = \frac{1}{9N^2}.$$
For the norm of the trigonometric term we get
$$|\pi\cot(\pi t + \pi i N)|
= \pi\left|\frac{e^{i\pi t - \pi N} + e^{-i\pi t + \pi N}}
{e^{i\pi t - \pi N} - e^{-i\pi t + \pi N}}\right|
\le \pi\left|\frac{e^{\pi N}+e^{-\pi N}}{e^{\pi N}-e^{-\pi N}}\right|
=\pi|\coth(\pi N)|.$$
There aren't any poles here either and this term is bounded above by $\pi\coth(\pi)$ because $N>1.$ This gives the following bound on the norm of the integral along $\Gamma_2:$
$$(2N+1) \times \frac{1}{9N^2} \times \pi\coth(\pi) \in \Theta(1/N)$$
and this integral also vanishes as $N$ goes to infinity as claimed.
The other two line segments can be bounded by the same technique.
