How to solve the following sum?
$$\sum _{n=1}^{\infty }\frac{1}{6n\left(n-1\right)\ +\ 1}$$
I was learning about Star Number then I saw on wikipedia that $\sum _{n=1}^{\infty }\frac{1}{S_n} = \dfrac{\pi}{2\sqrt3}\tan\left(\dfrac{\pi}{2\sqrt{3}}\right)$ where $S_n$ denotes the star number. I really don't know how to solve this sum.
I tried to approximate the infinite sum using integral,
$$\int_1^\infty \frac{1}{6n\left(n-1\right)\ +\ 1}\ dn$$
But this seems not working here.
I think this is quite relevant but I want to know any approach using some elementary techniques.
Best Answer
Ok, knowing the solution makes it simpler. Found $$ \pi \tan \pi x = 2x\sum_{n=0}^\infty \frac{1}{(n+\tfrac12)^2 - x^2}=:f(x) \tag 1$$ in Wiki, so let's work towards that: $$\begin{align} \sum_{n=1}^\infty\frac 1{6n(n-1)+1} &= \sum_{n=0}^\infty\frac 1{6n^2+6n+1}\\ &=\frac 16 \sum_{n=0}^\infty\frac 1{n^2+n+\frac14-\frac14+\frac16}\\ &=\frac 16 \sum_{n=0}^\infty\frac 1{(n+\frac12)^2-\frac 1{12}}\\ &=\frac16\cdot \frac{f(1/\sqrt{12})}{2\cdot1/\sqrt{12}}\\ &=\frac 1 {\sqrt{12}} \,f\Big(\frac 1 {\sqrt{12}}\Big) \end{align}$$
So what's left is to show the partial fraction expansion $(1)$, but that's a much more common and generic result from Euler, and proof should be easy to find. For example, here is Cotangent and the Herglotz Trick as PDF. Or here at MSE there is How did Euler prove the partial fraction expansion of the cotangent function?.