I have tried to evaluate the series $\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{n \log n}$. I get that it converges to $\frac{1}{2}$ but there must be a flaw in my reason since then result I get does not match that of Wolfram's Alpha.
Let me show you my work.
One good way to begin with is to recall the known Fourier series
\begin{equation} \sum_{n=1}^{\infty} \frac{\cos nx}{n^s} =\frac{1}{2} \bigg({\rm Li}_s \left ( e^{-ix} \right ) + {\rm Li}_s \left ( e^{ix} \right ) \bigg) \end{equation}
where ${\rm Li}_s$ is the polylogarithm of order $s$. Hence
\begin{equation} \sum_{n=2}^{\infty} \frac{\cos nx}{n^s} = - \cos x + \frac{1}{2} \bigg({\rm Li}_s \left ( e^{-ix} \right ) + {\rm Li}_s \left ( e^{ix} \right ) \bigg) \end{equation}
We differentiate with respect to $s$ , hence
\begin{equation} \sum_{n=2}^{\infty} \frac{\cos nx}{n^s \log n} = \frac{e^{ix}}{2} {\rm Li}_{s-1} \left ( e^{-ix} \right ) + \frac{e^{-ix}}{2} {\rm Li}_{s-1} \left ( e^{ix} \right ) \end{equation}
For $x=\pi$ and $s=1$ we have that $\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n} = \frac{1}{2}$.
Where have I gone wrong?
