How can we show that : $\sum\limits_{n=1}^{\infty}{H_{n,2}\over n^2}={7\over4}\cdot\zeta(4)$ Experimental on wolfram sum calculator yield 
$$\sum\limits_{n=1}^{\infty}{H_{n,2}\over n^2}={7\over4}\cdot\zeta(4)\tag1$$
Where $H_{n,2}=\sum\limits_{k=1}^{n}{1\over k^2}$
Can anyone show that (1) is correct?
My try:
The least I can do it to expand the series to see if any interesting pattern emerges
$$S={H_{1,2}\over 1^2}+{H_{2,2}\over 2^2}+{H_{3,2}\over 3^2}+{H_{4,2}\over 4^2}+\cdots$$
any way it is too long write it down but I manage to simplify to
${7\over4}\cdot\zeta(4)=\zeta(4)+{1\over 2^2}+{1\over 3^2}\left(1+{1\over2^2}\right)+{1\over4^2}\left(1+{1\over2^2}+{1\over3^2}\right)+{1\over5^2}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}\right)+\cdots$
${3\over4}\cdot\zeta(4)={1\over 2^2}+{1\over 3^2}\left(1+{1\over2^2}\right)+{1\over4^2}\left(1+{1\over2^2}+{1\over3^2}\right)+{1\over5^2}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}\right)+\cdots$
It becomes
$$\sum_{n=1}^{\infty}{H_{n,2}\over (n+1)^2}={3\over4}\cdot\zeta(4)$$
Or we can further simplify to
${3\over4}\cdot\zeta(4)=\zeta(2)-1+{1\over 3^2}\left({1\over2^2}\right)+{1\over4^2}\left({1\over2^2}+{1\over3^2}\right)+{1\over5^2}\left({1\over2^2}+{1\over3^2}+{1\over4^2}\right)+\cdots$
It can go another further than this.
 A: One may use multizeta values,
$$
\zeta(a,b)=\sum_{n_1> n_2>0}\frac{1}{n_1^{a} n_2^{b}}=\sum_{n=1}^\infty \frac{H_{n,b}}{(n+1)^a}
$$ then we have Euler's reflection formula
$$
\zeta(a,b)+\zeta(b,a)=\zeta(a)\zeta(b)-\zeta(a+b)
$$ here, $a=b=2$ gives
$$
\zeta(2,2)=\sum_{n=1}^\infty \frac{H_{n,2}}{(n+1)^2}=\frac{(\zeta(2))^2-\zeta(4)}2=\frac{\frac{\pi ^4}{36}-\frac{\pi ^4}{90}}2=\frac34\cdot \zeta(4)
$$ as wanted.
Going back to Euler's solution to the Basel problem, we have in fact (see here, pp. 6-7),

$$
\sum_{n=0}(-1)^n\zeta(\underbrace{2,2,\cdots,2,2}_{n})\cdot t^{2n}=\prod_{n=1}^\infty\left(1-\frac{t^2}{n^2} \right)=\frac{\sin (\pi t)}{\pi t}.
$$ 

yielding
$$
\zeta(\underbrace{2,2,\cdots,2,2}_{n})=\frac{\pi^{2n}}{(2n+1)!},\quad n\ge1.
$$
A: That is fairly easy with a simmetry argument given by the following Lemma:
$$ \sum_{n=1}^{N}f(n)\sum_{m=1}^{n} f(m) = \!\!\!\sum_{1\leq m\leq n\leq N}f(n)\,f(m)=\frac{1}{2}\left(\left(\sum_{n=1}^{N}f(n)\right)^2+\sum_{n=1}^{N}f(n)^2\right) \tag{1}$$
If we consider $f(n)=\frac{1}{n^2}$ and take the limit as $N\to +\infty$, we instantly get:
$$ \sum_{n\geq 1}\frac{H_n^{(2)}}{n^2} = \frac{\zeta(2)^2+\zeta(4)}{2} = \color{red}{\frac{7}{4}\,\zeta(4)}\tag{2}$$
as wanted.
A: Yes, it is correct. More generally, for $k>1$ and for $N>0$
$$H_{N,k}^2=2\sum_{n=1}^{N}{H_{n-1,k}\over n^k}+H_{N,2k}$$
and then by taking the limit we get
$$\zeta(k)^2=2\sum_{n=1}^{\infty}{H_{n-1,k}\over n^k}+\zeta(2k).$$
Hence
$$\sum_{n=1}^{\infty}{H_{n,k}\over n^k}=
\sum_{n=1}^{\infty}{H_{n-1,k}\over n^k}+\zeta(2k)=\frac{\zeta(k)^2+\zeta(2k)}{2}.$$
A: By Abel's summation we have 
$$\sum_{n=1}^N\frac{H_n^{(p)}}{n^q}=H_N^{(p)}H_N^{(q)}+H_N^{(p+q)}-\sum_{n=1}^N\frac{H_n^{(q)}}{n^p}$$
Set $p=q=2$ then let $N$ approach $\infty$ we get
$$2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}=\zeta^2(2)+\zeta(4)$$
Plugging $\zeta^2(2)=\frac52\zeta(4)$ we get 
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}=\frac74\zeta(4)$$
A: Different approach:
By Cauchy product we have 
$$\operatorname{Li}^2_2(x)=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac6{n^4}\right)x^n$$
Set $x=1$ and rearrange the terms we have 
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}=\frac12\zeta^2(2)+3\zeta(4)-2\sum_{n=1}^\infty\frac{H_n}{n^3}$$
Plug $\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$ and $\zeta^2(2)=\frac52\zeta(4)$, the result follows. 
