An Euler trigonometric sum

Prove that:

$$\sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n} \cos \frac{n \pi}{3}= -\frac{\pi^2}{36}$$

Solution

We might begin with the known generating function

$$\sum_{n=1}^{\infty} \mathcal{H}_n x^n = - \frac{\log (1-x)}{1-x} \Rightarrow \sum_{n=1}^{\infty} \mathcal{H}_n x^{n-1} = - \frac{\log(1-x)}{x (1-x)} \tag{1}$$

Integrating $(1)$ we get that:

$$\sum_{n=1}^{\infty} \frac{\mathcal{H}_n }{n}x^n = \frac{1}{2} \log^2 (1-x) + {\rm Li}_2(x)$$

where ${\rm Li}_2$ is the dilogarithmic function. We can also, easily, see that:

$$\mathfrak{Re}\left ( {\rm Li}_2 \left ( e^{ix} \right )  \right )= \sum_{n=1}^{\infty}\frac{\cos nx}{n^2} = \frac{x^2}{4}- \frac{\pi x}{2}+ \frac{\pi^2}{6} \tag{2}$$

Equation $(2)$ comes up easily if we integrate the known Fourier series $\sum \limits_{n=1}^{\infty} \frac{\sin nx}{n}=\frac{\pi-x}{2}$. Subbing $x=\frac{\pi}{3}$ at $(2)$ we get:

$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n}\cos \left ( \frac{n \pi}{3} \right ) &= \mathfrak{Re}\left ( \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n} \left ( e^{i\pi/3} \right )^n \right ) \\
 &= \frac{1}{2} \log^2 \left ( 1-e^{i\pi/3} \right )+ \mathfrak{Re}\left ( {\rm Li}_2 \left ( e^{i\pi/3} \right ) \right )\\
 &= \frac{1}{2} \log^2 \left ( 1-e^{i\pi/3} \right )+ \frac{\left ( \frac{\pi}{3} \right )^2}{4} - \pi \frac{\frac{\pi}{3}}{2} + \frac{\pi^2}{6}\\
 &= - \frac{\pi^2}{18}+ \frac{\pi^2}{36} \\
 &= - \frac{\pi^2}{36}
\end{aligned}$$