# Stuck on this tough limit involving polylogarithm

I need help proving this limit:

$$\lim_{x \to \infty}\left[% {2{\rm Li}_{2}\left(1-{\rm e}^{x}\right) \over x} -{x\,{\rm e}^{x} \over 1 - {\rm e}^{x}}\right] =0$$

Where the ${\rm Li}_{2}$ is polylogarithm function of order $2$.

The difficulty is because each component go to negative infinity, so we have to evaluate the limit of the whole expression. $$\mbox{I was able to show that}\quad\frac{\displaystyle{\frac{x\,{\rm e}^{x}}{1 - {\rm e}^{x}}}} {\displaystyle{\frac{2\,{\rm Li}_{2}\left(1 - {\rm e}^{x}\right)}{x}}}=1$$ but I have a hunch that the implication do not go that way. Please help quickly. My homework due soon. Thank you.

Using the identity \begin{align} Li_{2}(1-t) + Li_{2}\left(1 - \frac{1}{t}\right) = - \frac{1}{2} \ln^{2}(t) \end{align} with $t= e^{x}$ it is sen that \begin{align} Li_{2}(1-e^{x}) = - \frac{x^{2}}{2} - Li_{2}(1-e^{-x}). \end{align} Now the limit is \begin{align} \lim_{n \rightarrow \infty} \left\{ \frac{2}{x} \ Li_{2}(1-e^{x}) - \frac{x}{e^{-x}-1} \right\} &= \lim_{n \rightarrow \infty} \left\{ - \frac{2}{x} \ Li_{2}(1-e^{-x}) - x - \frac{x}{e^{-x} -1} \right\} \\ &= \lim_{n \rightarrow \infty} \left\{ - \frac{2}{x} \ Li_{2}(1-e^{-x}) - \frac{x}{1-e^{x}} \right\} \\ &= \lim_{n \rightarrow \infty} \left\{ - \frac{2}{x} \ Li_{2}(1-e^{-x}) - \frac{x}{1-(1 + x + \frac{x^{2}}{2} + \mathcal{O}(x^{3}))} \right\} \\ &= \lim_{n \rightarrow \infty} \left\{ - \frac{2}{x} \ Li_{2}(1-e^{-x}) + \frac{1}{\frac{x}{2} + \mathcal{O}(x^{3})} \right\} \\ &= 0. \end{align}
• The dilogarithm is the series \begin{align} Li_{2}(x) = \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \end{align} for which properties can be obtained. In the above case the limit is for $x \rightarrow infty$ the term $\frac{1}{x}$ dominates and makes the term zero. – Leucippus Dec 15 '14 at 7:10
• Instead of $n$ is should be the limit as $x \rightarrow \infty$ – Leucippus Dec 15 '14 at 7:11