How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ [closed]

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.

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

$$\begin{align} \sum_{n=1}^\infty\frac{1}{n(n+2)}\sum_{k=1}^{n+2}\frac1k&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+2)}\right)\\\\ &=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty\frac1{2k^2}+\frac12\\\\ &=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\frac{\pi^2}{12}+\frac12\\\\ &=\color{red}{\sum_{n=1}^\infty\left(\frac{1}{4n}-\frac{1}{4(n+1)}-\frac{1}{2(n+2)^2}\right)}\\\\ &+\color{blue}{\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}\right)}\\\\ &+\frac{\pi^2}{12}+\frac12\\\\ &=\color{blue}{1-\frac{\pi^2}{12}}+\color{red}{\frac14}+\frac{\pi^2}{12}+\frac12\\\\ &=\frac74 \end{align}$$

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}

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

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 $$\frac 1a\left(\frac{\pi^2}6+\left(H_a\right){}^2-\psi ^{(1)}(a+1)\right)$$