I learnt on www.pi314.net that $$\sum_{n=1}^{\infty}\frac{H_n}{n}\cos\left(\frac{n\pi}{3}\right)=-\frac{\pi^2}{36}$$
This result is hard to verify using Wolfram Alpha since the series converges very slowly.
I do not know how to prove this result.
I tried to rewrite the original series as
$$\sum_{k\geq1} \frac{1}{k}\sum_{n\geq k}\frac{\cos\left(\frac{n\pi}{3}\right)}{n}$$
I know that
$$\sum_{n\geq 1}\frac{\cos\left(\frac{n\pi}{3}\right)}{n}=0$$
But since $n$ starts from $k$, I cannot continue from here.
Any hint?
Best Answer
Perhaps it is in the extraction of the values you are having trouble with. If so, here is one rather (tedious and laborious) way to do it.
From Eq. (590) in the link, one has $$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \operatorname{Li}_2 (x) + \frac{1}{2} \ln^2 (1 - x).$$ Setting $x = e^{i\pi /3}$ we see that \begin{align} \sum_{n = 1}^\infty \frac{H_n}{n} \cos \left (\frac{n \pi}{3} \right ) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n} e^{i \pi n/3}\\ &= \operatorname{Re} \left [\operatorname{Li}_2 \left (e^{i\pi/3} \right ) + \frac{1}{2} \ln^2 \left (1 - e^{i \pi/3} \right ) \right ]. \end{align}
The log term can be disposed of immediately. Here $$\ln (1 - e^{i \pi/3}) = -\frac{i \pi}{3}.$$
For the dilogarithm term, $\operatorname{Re} \operatorname{Li}_2 (e^{i \theta})$ is an elementary function (for a reason why, see here) and means a value for this term can be found. What I present below is one way to find it (and there are no doubt other, simpler ways).
Starting with the series representation for the dilogarithm function, we have \begin{align} \operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] &= \operatorname{Re} \sum_{n = 1}^\infty \frac{e^{in \pi/3}}{n^2}\\ &= \sum_{n = 1}^\infty \frac{\cos \left (\frac{n \pi}{3} \right )}{n^2}\\ &= \sum_{\substack{n = 1\\n \in 6,12,\ldots}}^\infty \frac{1}{n^2} - \sum_{\substack{n = 1\\n \in 3,9,\ldots}}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{\substack{n = 1\\n \in 1,7,\ldots}}^\infty \frac{1}{n^2}\\ & \quad - \frac{1}{2} \sum_{\substack{n = 1\\n \in 2,8,\ldots}}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{\substack{n = 1\\n \in 5,11,\ldots}}^\infty \frac{1}{n^2} - \frac{1}{2} \sum_{\substack{n = 1\\n \in 4,10,\ldots}}^\infty \frac{1}{n^2}. \end{align} Terms in the series can be rearranged as the series absolutely converges. Shifting the indices as follows: $n \mapsto 6n$, $n \mapsto 6n + 3$, $n \mapsto 6n +1$, $n \mapsto 6n + 2$, $n \mapsto 6n + 5$, $n \mapsto 6n + 4$ leads to \begin{align} \operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] &= \frac{1}{36} \sum_{n =1}^\infty \frac{1}{n^2} - \frac{1}{36} \sum_{n = 0}^\infty \frac{1}{(n + 1/2)^2} + \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 1/6)^2}\\ & \quad - \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 1/3)^2} + \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 5/6)^2} - \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 2/3)^2}\\ &= \frac{1}{72} \left [2 \cdot \zeta (2) - 2 \psi^{(1)} \left (\frac{1}{2} \right ) + \psi^{(1)} \left (\frac{5}{6} \right ) + \psi^{(1)} \left (\frac{1}{6} \right ) - \psi^{(1)} \left (\frac{2}{3} \right ) - \psi^{(1)} \left (\frac{1}{3} \right ) \right ], \end{align} where $\psi^{(1)} (z)$ is the polygamma function (trigamma function).
Using the known values of $\zeta (2) = \pi^2/6$, $\psi^{(1)} (1/2) = \pi^2/2$, and making use of the reflection relation for the trigamma function of $$\psi^{(1)} (1 - z) + \psi^{(1)} (z) = \pi^2 \csc^2 (\pi z),$$ we see that $$\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] = \frac{1}{72} \left [2 \cdot \frac{\pi^2}{6} - 2 \cdot \frac{\pi^2}{2} + 4 \pi^2 - \frac{4 \pi^2}{3} \right ] = \frac{\pi^2}{36}.$$
Thus $$\sum_{n = 1}^\infty \frac{H_n}{n} \cos \left (\frac{n \pi}{3} \right ) = \frac{\pi^2}{36} - \frac{\pi^2}{18} = -\frac{\pi^2}{36},$$ as desired.