An asymptotic behavior of $\operatorname{Li}_{-n}(a)$ for $n\to\infty$ Suppose $a,b\in(0,1)$. I'm interested in comparison of an asymptotic behavior of $\operatorname{Li}_{-n}(a)$ and $\operatorname{Li}_{-n}(b)$ for $n\to\infty$.
Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case $\operatorname{Li}_{-n}\!\left(\tfrac12\right)$ for $n\ge1$ gives (up to a coefficient) a combinatorial sequence called Fubini numbers or ordered Bell numbers$^{[1]}$$\!^{[2]}$$\!^{[3]}$ (number of outcomes of a horse race provided that ties are possible). This sequence is known to have the following asymptotic behavior:
$$\operatorname{Li}_{-n}\!\left(\tfrac12\right)\sim\frac{n!}{\ln^{n+1}2}.\tag1$$
After some numerical experimentation I conjectured the following behavior:
$$\ln\!\left(\frac{\operatorname{Li}_{-n}(a)}{\operatorname{Li}_{-n}(b)}\right)=(n+1)\cdot\ln\!\left(\frac{\ln b}{\ln a}\right)+o\!\left(n^{-N}\right)\tag2$$
for arbitrarily large $N$ (so, the remainder term decays faster than any negative power of $n$). It looks like the remainder term is oscillating with exponentially decreasing amplitude, but I haven't yet found the exact exponent base or asymptotic oscillation frequency.
Could you suggest a proof of $(2)$ or further refinements of this formula?
 A: 
This is quite bizarre, but I believe that I can prove that your equation holds with the desired error term for any $$1>a,b>e^{-\pi}.$$ The asymptotic that I give for $L_{-n}(a)$ only holds when $1>a>e^{-\pi}$.

By definition, for $a<1$
 $$\text{Li}_{-n}(a)=\sum_{k=1}^{\infty}k^{n}a^{k},$$
 and we can approximate this sum by the integral $$\int_{0}^{\infty}x^{n}a^{x}dx=\int_{0}^{\infty}x^{n}e^{-x\log\left(\frac{1}{a}\right)}dx.$$
 Letting $x=\frac{u}{\log1/a}$
  the integral becomes $$\frac{1}{(\log\left(\frac{1}{a}\right))^{n+1}}\int_{0}^{\infty}u^{n}e^{-u}dx=\frac{\Gamma(n+1)}{\left(\log\left(\frac{1}{a}\right)\right)^{n+1}},$$
 and so $$\frac{\text{Li}_{-n}(a)}{\text{Li}_{-n}(b)}\approx\left(\frac{\log(1/b)}{\log(1/a)}\right)^{n+1}.$$
 Let's prove this with an exact error term by making the first step precise. Let $f_{a}(x)=x^{n}a^{x}$. Then since $$\frac{d^{k}}{dx^{k}}f_{a}(x)\biggr|_{x=0}=\frac{d^{k}}{dx^{k}}f_{a}(x)\biggr|_{x=\infty}=0$$
 for all $k<n$, Euler-Maclaurin summation up to $p=\frac{n}{2}$
  for even $n$
  implies that $$\left|\sum_{k=0}^{\infty}k^{n}a^{k}-\int_{0}^{\infty}x^{n}a^{x}dx\right| \le |R|
 \leq \frac{2\zeta(n)}{(2\pi)^{n}}\left(\int_{0}^{\infty}|f^{(n)}(x)|dx\right)
 \leq \frac{2\zeta(n)}{(2\pi)^{n}}\left(2^{n}\frac{\Gamma(n+1)}{\log(1/a)}\right)
 = O\left(\frac{1}{\pi^{n}}\frac{\Gamma(n+1)}{\log(1/a)}\right).$$
 Thus, $$\frac{\text{Li}_{-n}(a)}{\text{Li}_{-n}(b)}=\frac{\frac{\Gamma(n+1)}{\left(\log\left(\frac{1}{a}\right)\right)^{n+1}}+O\left(\frac{1}{\pi^{n}}\frac{\Gamma(n+1)}{\log(1/a)}\right)}{\frac{\Gamma(n+1)}{\left(\log\left(\frac{1}{b}\right)\right)^{n+1}}+O\left(\frac{1}{\pi^{n}}\frac{\Gamma(n+1)}{\log(1/b)}\right)}=\left(\frac{\log(1/b)}{\log(1/a)}\right)^{n+1}\left(\frac{1+O\left(\pi^{-n}\log(1/a)^{n}\right)}{1+O\left(\pi^{-n}\log(1/b)^{n}\right)}\right),$$
 and so $$\frac{\text{Li}_{-n}(a)}{\text{Li}_{-n}(b)}=\left(\frac{\log(1/b)}{\log(1/a)}\right)^{n+1}\left(1+O\left(\pi^{-n}\log\left(\frac{1}{\min(a,b)}\right)^{n}\right)\right),$$
 and using the fact that $\log(1+x)=x+O(x^{2})$
 , this implies implies that $$\log\left(\frac{\text{Li}_{-n}(a)}{\text{Li}_{-n}(b)}\right)=(n+1)\log\left(\frac{\log(1/b)}{\log(1/a)}\right)+O\left(\pi^{-n}\log\left(\frac{1}{\min(a,b)}\right)^{n}\right).$$
 Now, this error term is nontrivial only when $$1>a,b>e^{-\pi},$$ and in this case it is $o(n^{-N})$ for any fixed capital $N$. 
