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.
 A: 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}
A: recall: $\displaystyle H_a = \int_0^1 \frac{1-x^a}{1-x}\,\mathrm{d}x$, and integration by parts once we have
$$\int_0^1 x^{a-1} \ln (1-x)\,\mathrm{d}x = -\frac{H_a}{a}$$
Thus,
\begin{align*}&\sum\limits_{n=1}^{\infty} \frac{H_{n+a}}{n(n+a)} = -\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^{1} x^{n+a-1} \ln (1-x)\,\mathrm{d}x= \int_0^{1} x^{a-1} \ln^{2}(1-x)\,\mathrm{d}x\\&= \lim\limits_{b \to 1}\frac{\partial^2}{\partial b^2}\int_0^{1} x^{a-1}(1-x)^{b-1}\,\mathrm{d}x= \lim\limits_{b \to 1} \frac{\partial^2}{\partial b^2} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\\&= \lim\limits_{b \to 1}\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left((\psi_0(b) - \psi_0(a+b))^2 + \psi_1(b) - \psi_1(a+b)\right)\\&= \frac{1}{a}\left((\gamma + \psi_0(a+1))^2 + \frac{\pi^2}{6} - \psi_1(a+1)\right)\end{align*}
Hence put $a=2$ we get
$$\sum\limits_{n=1}^{\infty} \frac{H_{n+2}}{n(n+2)} =\frac{1}{2}\left [ \left ( \gamma +\underset{\psi _{0}\left ( 3 \right )}{\underbrace{\frac{3}{2}-\gamma}}  \right )^{2}+\frac{\pi ^{2}}{6}-\underset{\psi _{1}\left ( 3 \right )}{\underbrace{\left (\frac{\pi ^{2}}{6}-\frac{5}{4}  \right )} }\right ]=\frac{7}{4}$$
A: This is not an answer but it is too long for  comment.
Just out of curiosity and playing with a CAS,
$$S_p=\sum_{n=1}^{ p } \frac{H_{n+2}}{n(n+2)}=\frac{-2 \left(2 p^2+9 p+9\right) H_{p+3}+(7 p^3+38 p^2+64 p+33)}{4 (p+1) (p+2)
   (p+3)}$$ Using the asymptotics of harmonic numbers $$S_p=\frac{7}{4}-\frac{\log \left(p\right)+\gamma
   +1}{p}+O\left(\frac{1}{p^2}\right)$$
Similarly, considering $$T_p^{(a)}=\sum_{n=1}^{ p } \frac{H_{n+a}}{n(n+a)}$$ we get similar forms 
$$T_p^{(1)}=\frac{-(p+2) H_{p+2}+(2 p^2+5 p+3)}{(p+1) (p+2)}$$
$$T_p^{(3)}=\frac{-18 \left(3 p^3+24 p^2+59 p+44\right) H_{p+4}+(85 p^4+796 p^3+2624 p^2+3563
   p+1650)}{54 (p+1) (p+2) (p+3) (p+4)}$$ for which the asymptotics are
$$T_p^{(1)}=2-\frac{\log \left(p\right)+\gamma
   +1}{p}+O\left(\frac{1}{p^2}\right)$$
$$T_p^{(2)}=\frac{7}{4}-\frac{\log \left(p\right)+\gamma
   +1}{p}+O\left(\frac{1}{p^2}\right)$$
$$T_p^{(3)}=\frac{85}{54}-\frac{\log \left(p\right)+\gamma
   +1}{p}+O\left(\frac{1}{p^2}\right)$$
$$T_p^{(4)}=\frac{415}{288}-\frac{\log \left(p\right)+\gamma
   +1}{p}+O\left(\frac{1}{p^2}\right)$$
$$T_p^{(5)}=\frac{12019}{9000}-\frac{\log \left(p\right)+\gamma
   +1}{p}+O\left(\frac{1}{p^2}\right)$$ For sure, all limits correspond to Renascence_5's answer which can also write $$\frac 1a\left(\frac{\pi^2}6+\left(H_a\right){}^2-\psi ^{(1)}(a+1)\right)$$
A: 
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
$$\begin{align}
\sum_{n=1}^\infty\frac{1}{n(n+2)}\sum_{k=1}^{n+2}\frac1k&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+2)}\right)\\\\
&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty\frac1{2k^2}+\frac12\\\\
&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\frac{\pi^2}{12}+\frac12\\\\
&=\color{red}{\sum_{n=1}^\infty\left(\frac{1}{4n}-\frac{1}{4(n+1)}-\frac{1}{2(n+2)^2}\right)}\\\\
&+\color{blue}{\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}\right)}\\\\
&+\frac{\pi^2}{12}+\frac12\\\\
&=\color{blue}{1-\frac{\pi^2}{12}}+\color{red}{\frac14}+\frac{\pi^2}{12}+\frac12\\\\
&=\frac74
\end{align}$$
