I have been playing around with dilogarithm and considered the following sum.
$$\sum_{n=1}^{k} {\text{Li}_{2} {(\omega^n)}}$$
Where $\text{Li}_2{(x)}$ is the Dilogarithm and $\omega$ is the k-th root of unity. I conjecture it equals $\zeta(2)/k$, but I am not sure how to tackle this. Also, I think the limit as k approaches infinity seems interesting.
I'm assuming that $\omega$ is a primitive $k$-th root of unity.
Then $$\sum_{n=1}^k\text{Li}_2(\omega^n)=\sum_{n=1}^k\sum_{j=1}^\infty \frac{\omega^{nj}}{j^2}=\sum_{j=1}^\infty\frac{1}{j^2}\sum_{n=1}^k \omega^{nj}$$
If $k\mid j$ then $\omega^{j}=1$ and $\sum_{n=1}^k \omega^{nj}=k$, and if $k\not\mid j$ then $\omega^{j}\ne1$, so we can calculate the geometric sum $$\sum_{n=1}^k \omega^{nj}=\omega^j\frac{\omega^{jk}-1}{\omega^j-1}=0$$
Then $$=\sum_{j=1}^\infty\frac{1}{j^2}\sum_{n=1}^k \omega^{nj}=\sum_{j=1\\ k|j}^\infty\frac{1}{j^2}k=\sum_{l=1}^\infty\frac{1}{(lk)^2}k = \frac{\zeta (2)}{k}$$