<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] 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
$$\sum\limits_{n=1}^{\infty} \frac{H_{n+2}}{n(n+2)} =\frac{1}{2}\left [ \left ( \gamma +\underset{\psi _{0}\left ( 3 \right )}{\underbrace{\frac{3}{2}-\gamma}}  \right )^{2}+\frac{\pi ^{2}}{6}-\underset{\psi _{1}\left ( 3 \right )}{\underbrace{\left (\frac{\pi ^{2}}{6}-\frac{5}{4}  \right )} }\right ]=\frac{7}{4}$$
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