How prove this $\sum_{n=1}^{\infty}\frac{\zeta_{2}}{n^4}=\zeta^2(3)-\frac{1}{3}\zeta(6)$ show that
$$\sum_{n=1}^{\infty}\dfrac{\zeta_{2}}{n^4}=\zeta^2(3)-\dfrac{1}{3}\zeta(6)$$
where
$$\zeta_{m}=\sum_{k=1}^{n}\dfrac{1}{k^m},\zeta(m)=\sum_{k=1}^{\infty}\dfrac{1}{k^m}$$
is true？
because This result is my frend tell me.
This problem have someone research it?Thank you
my some idea:
$$\zeta^3(3)=\left(\sum_{n=0}^{\infty}\dfrac{1}{(n+1)^3}\right)^2=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\dfrac{1}{(k+1)^3(n-k+1)^3}$$
and use
$$\dfrac{1}{(k+1)(n-k+1)}=\dfrac{1}{n+2}\left(\dfrac{1}{k+1}+\dfrac{1}{n-k+1}\right)$$
and $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
and
$$\sum_{n=1}^{\infty}\dfrac{H_{n}}{(n+1)^5}=\dfrac{1}{2}\left(5\zeta(6)-2\zeta(2)\zeta(4)-\zeta^2(3)\right)$$
But is very ugly, someone  have other nice methods? Thank you . 
 A: Consider $$f(z) = \frac{\Big(\psi_{1}(-z) \Big)^{2}}{z^{\color{red}{3}}}$$
(where $\psi_{1}(z)$ is the trigamma function) and integrate around a circle centered at the origin of complex plane of radius $N + \frac{1}{2}$ where $N$ is a positive integer.
Then $$\lim_{N \to \infty } \int_{|z| = N+\frac{1}{2}} f(z) \ dz = 0 = 2 \pi i \left(\text{Res}[f(z),0] + \sum_{n=1}^{\infty} \text{Res}[f(z),n] \right) .$$
For $ n \ge 1 $,
$$ \begin{align} f(z) &= \frac{1}{z^{3}} \Bigg[\frac{1}{(z-n)^{2}}+ \Big(H_{n}^{(2)} + \zeta(2) \Big) - 2 \Big(H_{n}^{(3)}-\zeta(3)\Big) (z-n) + \mathcal{O}\Big((z-n)^{2}\Big) \Bigg]^{2} \\ &= \frac{1}{z^{3}} \Bigg(\frac{1}{(z-n)^{4}} + \frac{2 H_{n}^{(2)}+2 \zeta(2)}{(z-n)^{2}} - \frac{4H_{n}^{(3)} -4 \zeta(3)}{z-n} + \mathcal{O}(1) \Bigg) \end{align}$$ 
where I'm using the notation $$H_{n}^{(m)} = \sum_{k=1}^{n} \frac{1}{k^{m}} .$$
Therefore,
$$ \begin{align} \text{Res} [f(z),n] &= \text{Res} \Bigg[ \frac{1}{z^{3}} \frac{1}{(z-n)^{4}},n \Bigg] + \text{Res} \Bigg[ \frac{1}{z^{3}} \frac{2 H_{n}^{(2)} + 2 \zeta(2)}{(z-n)^{2}},n\Bigg] + \text{Res} \Bigg[ \frac{4 \zeta(3) - 4 H_{n}^{(3)}}{z-n},n\Bigg] \\ &= -\frac{10}{n^{6}} - \frac{6 H_{n}^{(2)}}{n^{4}} - \frac{6 \zeta(2)}{n^{4}} + \frac{4 \zeta(3)}{n^{3}}- \frac{4 H_{n}^{(3)}}{n^{3}} . \end{align}$$
And at the origin,
$$ \begin{align} f(z) &= \frac{1}{z^{3}} \Big( \frac{1}{z^{2}} + \zeta(2) + 2 \zeta(3) z + 3 \zeta(4) z^{2} + 4 \zeta(5) z^{3} + 5 \zeta(6)z^{4} + \mathcal{O}(z^{5}) \Big)^{2} \\ &= \frac{1}{n^{7}} + \frac{2 \zeta(3)}{z^{5}} + \frac{2 \zeta(3)}{z^{4}} + \frac{6 \zeta(4) + \zeta^{2}(z)}{z^{3}} + \frac{9 \zeta(5) + 4 \zeta(2) \zeta(3)}{z^{2}} \\ &+ \frac{10 \zeta(6) + 6 \zeta(2) \zeta(4)+ 4 \zeta^{2}(3)}{z} + \mathcal{O}(1) . \end{align}$$
So
$$ \text{Res}[f(z),0] = 10 \zeta(6) + 6 \zeta(2) \zeta(4)+ 4 \zeta^{2}(3) .$$
Summing up all the residues we have 
$$ -10 \zeta(6) - 6 \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} - 6 \zeta(2) \zeta(4) + 4 \zeta^{2}(3) - 4 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{3}}  + 10 \zeta(6) + 6 \zeta(2) \zeta(4) + 4 \zeta^{2}(3) =0$$
$$ \implies 6 \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} = 8 \zeta^{2}(3) - 4 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{3}} .$$
In general, $$\sum_{n=1}^{\infty} \frac{H_{n}^{(m)}}{n^{m}} = \frac{\zeta^{2}(m) + \zeta(2m)}{2} .$$
Therefore,
$$ \begin{align} \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} &= \frac{1}{6} \Bigg( 8 \zeta^{2}(3)- 4 \Big( \frac{\zeta^{2}(3)+\zeta(6)}{2} \Big) \Bigg) \\ &= \zeta^{2}(3) - \frac{\zeta(6)}{3} . \end{align}$$
A: I introduced an integral representation for a more general case in a previous problem

$$ B(p,q) = \sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q}=\frac{(-1)^q}{\Gamma(q)}\int_{0}^{1}\!{\frac {\left(\ln\left(u\right)\right)^{q-1}{Li_{p}(u)} }{ u\left( u-1 \right)}}{du} $$

in which your sum can have the form

$$ B(2,4) = \sum_{k=1}^{\infty} \dfrac{H_k^{(2)}}{k^4}=\frac{1}{\Gamma(4)}\int_{0}^{1}\!{\frac {\left(\ln\left(u\right)\right)^{3}{Li_{2}(u)} }{ u\left( u-1 \right)}}{du}. $$

Try to manipolate this integral. See a related problem.
