Let $H_{n}$ be the

nth harmonic number defined by $ H_{n} := \sum_{k=1}^{n} \frac{1}{k}$.How would you prove that

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$

Simply replacing $H_{n}$ with $\sum_{k=1}^{n} \frac{1}{k}$ does not seem like a good starting point. Perhaps another representation of the

nth harmonic number would be more useful.

**Answer**

The Euler sum $\sum_{n=1}^{\infty} \frac{H_{n}}{n^{q}}$, where $q$ is an odd positive integer greater than $1$, can also be evaluated using this approach. See here.

Using the integration representation $$H_{n} = \int_{0}^{1} \frac{1-t^{n}}{1-t} \, dt \ ,$$

we have

$$ \begin{align} \sum_{n=1}^{\infty} \frac{H_{n}}{n^{3}} &= \sum_{n=1}^{\infty} \frac{1}{n^{3}} \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt \\ &= \int_{0}^{1} \frac{1}{1-t} \sum_{n=1}^{\infty} \frac{1-t^{n}}{n^{3}} \\ &=\int_{0}^{1} \frac{\zeta(3) – \text{Li}_{3}(t)}{1-t} \ dt \tag{1} \\ &=- \Big(\zeta(3)-\text{Li}_{3}(t)\Big) \ln(1-t) \Bigg|_{0}^{1} – \int^{1}_{0} \frac{ \text{Li}_{2}(t) \log(1-t)}{t} \ dt \\ &= -\int_{0}^{1} \frac{ \text{Li}_{2}(t) \log(1-t)}{t} \ dt \\ &= \int_{0}^{1} \text{Li}_{2}(t) \, d \big(\text{Li}_{2}(t)\big) \\ &= \frac{\big(\text{Li}_{2}(t)\big)^{2}}{2} \Bigg|^{1}_{0} \\ &= \frac{\zeta^{2}(2)}{2} \\ &= \frac{\pi^{4}}{72}. \end{align}$$

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**Attribution***Source : Link , Question Author : gauss115 , Answer Author : Community*