How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I have never done these questions before. If possible, please give a way out without using polylogarithmic functions or other non-elementary functions.

Actually the calculation for this sum is very simple and what we need is the sum of telescopic series. In fact
\begin{align}
\sum_{n=1}^\infty\frac{H_{n+2}}{n(n+2)}&=\frac12\sum_{n=1}^\infty H_{n+2}\left(\frac{1}{n}-\frac1{n+2}\right)\\\\
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{1}{n}\left(H_{n}+\frac{1}{n+1}+\frac{1}{n+2}\right)-\frac{H_{n+2}}{n+2}\right)\\\\ 
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{H_n}{n}+\frac{1}{n(n+1)}+\frac{1}{n(n+2)}-\frac{H_{n+2}}{n+2}\right)\\\\
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{H_{n+2}}{n+2}\right)+\frac12\sum_{n=1}^\infty\left(\frac{1}{n(n+1)}+\frac{1}{n(n+2)}\right)\\\\
&=\frac12\left(H_1+\frac{H_2}2\right)+\frac78\\\\
&=\frac{7}{4}.
\end{align}
