infinite series involving harmonic numbers and zeta

Solution 1:

$$\sum_{n=1}^{+\infty} \frac{H_{n}}{n^{3}} = \sum_{n=1}^{+\infty} \frac{1}{n^{3}} \sum_{m=1}^{+\infty} \left( \frac{1}{m} - \frac{1}{m+n}\right) = \sum_{n=1}^{+\infty} \frac{1}{n^{3}} \sum_{m=1}^{+\infty} \frac{n}{m(m+n)} = \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{m}{m^2 n^2 (m+n)} = \frac{1}{2} \left(\sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{m}{m^2 n^2(m+n)} + \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{n}{m^2 n^2(m+n)} \right) = \frac{1}{2} \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{1}{m^2 n^2} = \frac{1}{2} \zeta(2)^2 = \frac{1}{2} \left(\frac{\pi^{2}}{6}\right)^2 = \frac{\pi^{4}}{72} = \frac{5}{4} \zeta(4) $$

Solution 2:

See here: (father and son)

On An Intriguing Integral and Some Series Related to $\zeta(4)$ - David Borwein and Jonathan M. Borwein

Enjoy

Solution 3:

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:

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=-\int_{0}^{1}\frac{Li_{2}(1-x)\ln(x)}{1-x}dx$

$=\frac{1}{2}(Li_{2}(1-x))^{2} |_{0}^{1}$

$=\frac{1}{2}(Li_{2}(1))^{2}=\frac{1}{2}\left(\frac{{\pi}^{2}}{6}\right)^{2}$

$=\frac{{\pi}^{4}}{72}$.