infinite series involving harmonic numbers and zeta I ran across a fun looking series and am wondering how to tackle it. 
$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=\frac{{\pi}^{4}}{72}.$$
One idea I had was to use the digamma and the fact that 
$$\sum_{k=1}^{n}\frac{1}{k}=\int_{0}^{1}\frac{1-t^{n}}{1-t}dt=\psi(n+1)+\gamma.$$
Along with the identity $\psi(n+1)=\psi(n)+\frac{1}{n}$,  I managed to get it into the form
$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=\gamma\zeta(3)+\zeta(4)+\sum_{n=1}^{\infty}\frac{\psi(n)}{n^{3}}.$$
This would mean that $$\sum_{n=1}^{\infty}\frac{\psi(n)}{n^{3}}=\frac{{\pi}^{4}}{360}-\gamma\zeta(3).$$  Which, according to Maple, it does.  But, how to show it?.  If possible.
I also started with $\frac{-\ln(1-x)}{x(1-x)}=\sum_{n=1}^{\infty}H_{n}x^{n-1}$.
Then divided by x and differentiated several times.  This lead to some interesting, but albeit, tough integrals involving the dilog:
$$-\int\frac{\ln(1-x)}{x(1-x)}dx=Li_{2}(x)+\frac{\ln^{2}(1-x)}{2}=\sum_{n=1}^{\infty}\frac{H_{n}x^{n}}{n}.$$
Doing this again and again lead to some integrals that appeared to be going in the right direction.
$$\int_{0}^{1}\frac{Li_{3}(x)}{x}dx=\frac{{\pi}^{4}}{90}$$
$$-\int_{0}^{1}\frac{\ln^{2}(1-x)\ln(x)}{2x}dx=\frac{{\pi}^{4}}{360}$$
$$-\int_{0}^{1}\frac{\ln(1-x)Li_{2}(1-x)}{x}dx=\frac{{\pi}^{4}}{72}$$
But, what would be a good approach for this one? I would like to find out how to evaluate 
$$\sum_{n=1}^{\infty}\frac{\psi(n)}{n^{3}}=\frac{{\pi}^{4}}{360}-\gamma\zeta(3)$$
 if possible, but any methods would be appreciated and nice. 
Thanks a bunch. 
 A: See here: (father and son)
On An Intriguing Integral and Some Series Related to   $\zeta(4)$  - David Borwein and Jonathan M. Borwein 
Enjoy
A: I appreciate all of the input. 
I thought I would come back and post something I managed to come up with.
This is kind of based on the methods in my first post using the dilog.
I started by using the identity $-n\int_{0}^{1}(1-x)^{n-1}\ln(x)dx=-\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k}}{k}=H_{n}$. 
Then, $\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=-\sum_{n=1}^{\infty}\frac{1}{n^{2}}\int_{0}^{1}(1-x)^{n-1}\ln(x)dx$
$=-\int_{0}^{1}\sum_{n=1}^{\infty}\frac{(1-x)^{n-1}\ln(x)}{n^{2}}dx$
Using the definition of the dilog, $Li_{2}(1-x)=\sum_{n=1}^{\infty}\frac{(1-x)^{n}}{n^{2}}$, I got:
$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=-\int_{0}^{1}\frac{Li_{2}(1-x)\ln(x)}{1-x}dx$
$=\frac{1}{2}(Li_{2}(1-x))^{2} |_{0}^{1}$
$=\frac{1}{2}(Li_{2}(1))^{2}=\frac{1}{2}\left(\frac{{\pi}^{2}}{6}\right)^{2}$
$=\frac{{\pi}^{4}}{72}$.
A: $$\sum_{n=1}^{+\infty} \frac{H_{n}}{n^{3}} =
\sum_{n=1}^{+\infty} \frac{1}{n^{3}} \sum_{m=1}^{+\infty} \left( \frac{1}{m} - \frac{1}{m+n}\right)
= \sum_{n=1}^{+\infty} \frac{1}{n^{3}} \sum_{m=1}^{+\infty} \frac{n}{m(m+n)}
= \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{m}{m^2 n^2 (m+n)}
= \frac{1}{2} \left(\sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{m}{m^2 n^2(m+n)} + \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{n}{m^2 n^2(m+n)} \right) 
= \frac{1}{2} \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{1}{m^2 n^2}
= \frac{1}{2} \zeta(2)^2
= \frac{1}{2} \left(\frac{\pi^{2}}{6}\right)^2 
= \frac{\pi^{4}}{72}
= \frac{5}{4} \zeta(4) $$
