# How to show $\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=2\zeta (3)$? [duplicate]

How to show this equation is true.

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

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

• Actually, I almost regret to have voted to close this as a duplicate, "no personal input" would have been more fitting. To wit, the proof on the other page for $\sum\limits_n\frac{H_n}{n^q}$ makes some twists and turns, I feel, so much so that I got lost trying to follow it and that I began to rewrite it when $q=2$, that is, in the present case. The proof that resulted from this is completely elementary, rather short, and it allows to show some nice tools at work, such as symmetrization of discrete sums. – Did May 21 '14 at 20:06

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}
• please explain how you got from \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} – Amad27 Nov 9 '14 at 11:30