Maximum/minimum of a special function I was given a function $f(x)=\mbox{Li}_{-n}(x)$, where Li is the polylogarithm of order $-n$ ($n>0\in\mathbb{N}$) and $x\in(-\infty,0)$. The function in this domain is bounded and has some extremes. I would like o obtain analytically the extreme (maximum/minimum) of this function. I have tried to use the property $$\frac{d}{dx}\mbox{Li}_{-n}(x)=\frac{1}{x}\mbox{Li}_{-n-1}(x)=0$$
but I got stacked, since I don't know how to solve the zeros of the polylogarithm.
Note: Just in case it is useful, I have checked with numerics that the maximum scales as $\sim n!$, but I would like to obtain it more analytically.
 A: We have:
$$\begin{eqnarray*} \text{Li}_{-k}(x) = \sum_{n\geq 1}n^k x^n = \left.\frac{d^k}{dt^k}\sum_{n\geq 1}e^{nt} x^n\right|_{t=0}&=&\sum_{j=0}^{k}j!{k+1\brace j+1}\left(\frac{x}{1-x}\right)^{j+1}\\[0.2cm]&=&\frac{1}{(1-x)^{k+1}}\sum_{j=0}^{k-1}\left\langle k\atop j\right\rangle x^{k-j}\end{eqnarray*} $$
as a consequence of Worpitzky's identity, too. Since $f(x)=\frac{x}{1-x}$ is a bijective function from $\mathbb{R}^-$ to $(-1,0)$, in order to compute the stationary points of $\text{Li}_{-k}(x)$ over $\mathbb{R}^-$ it is enough to compute the stationary points of:
$$ g_k(x) = \sum_{j=0}^{k}j!{k+1\brace j+1}x^{j+1}=\text{Li}_{-k}\left(\frac{x}{x+1}\right) $$
over $(-1,0)$. Dobinski's formula gives:
$$B_k(z)=\sum_{j=0}^{k}{k \brace j}z^j $$
where the LHS  is a Bell polynomial. So we have:
$$ B_{k+1}(xz) = \sum_{j=0}^{k+1}{k+1\brace j+1} z^{j+1} x^{j+1} $$
and:
$$ g_k(x) = -(k+1)!x^{k+2}+\int_{0}^{+\infty}\frac{B_{k+1}(xz)}{ze^{z}}\,dz.$$
It follows that a more accurate approximation for $\sup_{u<0}\left|\text{Li}_{-k}(u)\right|$ should be given in terms of Bell numbers and complementary Bell numbers. Work in progress.
