summation - How to show $sum_{n=1}^{infty}frac{H_{n}}{n^{2}}=2zeta (3)$?

How to show this equation is true.

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=2\zeta (3)$$

where $H_{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$

Answer

It can be shown that
\begin{align}
H_{n} = \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt.
\end{align}
Using this integral form of the Harmonic numbers the series in question becomes
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n^{2}} \cdot \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt \\

&= \int_{0}^{1} \left[ \zeta(2) - Li_{2}(t) \right] \frac{dt}{1-t},
\end{align}
where $Li_{2}(x)$ is the dilogarithm. Now using the integral
\begin{align}
\int \frac{Li_{2}(t)}{1-t} \ dt = 2 Li_{3}(1-t) - 2 Li_{2}(1-t) \ \ln(1-t) - Li_{2}(t) \ \ln(1-t) - \ln(t) \ \ln^{2}(1-t)
\end{align}
it is seen that, with the use of $Li_{m}(1) = \zeta(m)$, $Li_{m}(0) = 0$, $\ln(1) = 0$,
\begin{align}
S &= \zeta(2) [ - \ln(1-t)]_{0}^{1} + \zeta(2) \ln(0) + 2 Li_{3}(1) \\
&= - \zeta(2) \ln(o) + \zeta(2) \ln(0) + 2 Li_{3}(1) \\

&= 2 Li_{3}(1)
\end{align}
which yields
\begin{align}
\sum_{n=1}^{\infty} \frac{H_{n}}{n^{2}} = 2 \zeta(3).
\end{align}