Value of the series $\sum_{n=1}^{\infty}\big (\frac{H_n}{\binom{3n}{n}}\big)^2$

I want to find the value of following series

$$\sum_{n=1}^{\infty} \left(\frac{H_n}{\binom{3n}{n}}\right)^2\tag{1}$$ where $H_n = 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$.

I know that $$\sum_{n=1}^{\infty} H_nx^n = -\frac{\log(1-x)}{1-x}. \tag{2}$$

and how to find

$$\sum_{n=1}^{\infty} \frac{x^n}{\binom{3n}{n}}\tag{3}$$ by using the integral from the Beta function; but I have no idea to find the value of $(1)$.

Any tips would be greatly appreciated. Step by step solutions would be hugely appreciated, thank you very much.

We have: $$\begin{eqnarray*} \frac{1}{\binom{3n}{n}}=\frac{n!(2n)!}{(3n)!}=\frac{\Gamma(n+1)\Gamma(2n+1)}{\Gamma(3n+1)}&=&(3n+1)\cdot B(n+1,2n+1) \\&=&(3n+1)\int_{0}^{1}\left(x^2(1-x)\right)^n\,dx\end{eqnarray*}$$ hence if we consider: $$\frac{3 z-\log(1-z)-2 z\log(1-z)}{(1-z)^2} = \sum_{n\geq 1}(3n+1)H_n z^n$$ we have: $$\begin{eqnarray*}f(t)&=&\sum_{n\geq 1}\frac{H_n}{\binom{3n}{n}}t^n\\ &=& \int_{0}^{1}\frac{3 tx^2(1-x)-\log(1-tx^2(1-x))-2 tx^2(1-x)\log(1-tx^2(1-x))}{(1-tx^2(1-x))^2}\,dx\end{eqnarray*}$$
and by Parseval's identity: $$\sum_{n\geq 1}\left(\frac{H_n}{\binom{3n}{n}}\right)^2 = \frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta})\cdot f(e^{-i\theta})\,d\theta$$ the original series turns into a (quite convoluted, but manageable) logarithmic integral.
• How to find $$\sum_{n>=1} (3n+1)H_nz^n$$ Thank you sir. – James Thomas Jun 2 '16 at 18:21
• @JamesThomas: by starting from your $(2)$ and computing its derivative multiplied by $x$. – Jack D'Aurizio Jun 2 '16 at 18:23
• I know that how to find $$\sum_{n>=1} (n+k)H_nz^n$$ but I have no idea for 3n+1. – James Thomas Jun 2 '16 at 18:29