<s>[INST] 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.
 [/INST] 
I thought it would be instructive to present a way forward that relies on elementary analysis, including knowledge of the sum $\sum_{k=1}^\infty \frac1{k^2}=\pi^2/6$, partial fraction expansion, and telescoping series.  It is to that end we proceed.


The Harmonic number, $H_{n+2}$, can be written as 
$$H_{n+2}=\sum_{k=1}^{n+2}\frac1k=\frac{1}{n+2}+\frac{1}{n+1}+\sum_{k=1}^n \frac1k$$
Therefore, we have
$$\begin{align}
\sum_{n=1}^\infty\frac{1}{n(n+2)}\sum_{k=1}^{n+2}\frac1k&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+2)}\right)\\\\
&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty\frac1{2k^2}+\frac12\\\\
&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\frac{\pi^2}{12}+\frac12\\\\
&=\color{red}{\sum_{n=1}^\infty\left(\frac{1}{4n}-\frac{1}{4(n+1)}-\frac{1}{2(n+2)^2}\right)}\\\\
&+\color{blue}{\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}\right)}\\\\
&+\frac{\pi^2}{12}+\frac12\\\\
&=\color{blue}{1-\frac{\pi^2}{12}}+\color{red}{\frac14}+\frac{\pi^2}{12}+\frac12\\\\
&=\frac74
\end{align}$$
</s>