 # "An asymptotic behavior of Li−n(a) for n-> inf Suppose a,b in (0,1). I'm interested in comparison of an asymptotic behavior of Li−n(a) and Li−n(b) for n-> inf. Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case Li−n(12) for n>=1 gives (up to a coefficient) a combinatorial sequence called Fubini numbers or orderded Bell numbers (number of outcomes of a horse race provided that ties are possible). mydaruma25 2022-09-17 Answered
An asymptotic behavior of ${\mathrm{Li}}_{-n}\left(a\right)$ for $n\to \mathrm{\infty }$
Suppose $a,b\in \left(0,1\right)$. I'm interested in comparison of an asymptotic behavior of ${\mathrm{Li}}_{-n}\left(a\right)$ and ${\mathrm{Li}}_{-n}\left(b\right)$ for $n\to \mathrm{\infty }$.
Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case ${\mathrm{Li}}_{-n}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{2}\right)$ for $n\ge 1$ gives (up to a coefficient) a combinatorial sequence called Fubini numbers or orderded Bell numbers (number of outcomes of a horse race provided that ties are possible). This sequence is known to have the following asymptotic behavior:
$\begin{array}{}\text{(1)}& {\mathrm{Li}}_{-n}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{2}\right)\sim \frac{n!}{{\mathrm{ln}}^{n+1}2}.\end{array}$
After some numerical exprerimentation I conjectured the following behavior:
$\begin{array}{}\text{(2)}& \mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{{\mathrm{Li}}_{-n}\left(a\right)}{{\mathrm{Li}}_{-n}\left(b\right)}\right)=\left(n+1\right)\cdot \mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{\mathrm{ln}b}{\mathrm{ln}a}\right)+o\phantom{\rule{negativethinmathspace}{0ex}}\left({n}^{-N}\right)\end{array}$
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?
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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 ${L}_{-n}\left(a\right)$ only holds when $1>a>{e}^{-\pi }$
By definition, for a<1
${\text{Li}}_{-n}\left(a\right)=\sum _{k=1}^{\mathrm{\infty }}{k}^{n}{a}^{k},$
and we can approximate this sum by the integral
${\int }_{0}^{\mathrm{\infty }}{x}^{n}{a}^{x}dx={\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x\mathrm{log}\left(\frac{1}{a}\right)}dx.$
Letting $x=\frac{u}{\mathrm{log}1/a}$ the integral becomes
$\frac{1}{\left(\mathrm{log}\left(\frac{1}{a}\right){\right)}^{n+1}}{\int }_{0}^{\mathrm{\infty }}{u}^{n}{e}^{-u}dx=\frac{\mathrm{\Gamma }\left(n+1\right)}{{\left(\mathrm{log}\left(\frac{1}{a}\right)\right)}^{n+1}},$
and so
$\frac{{\text{Li}}_{-n}\left(a\right)}{{\text{Li}}_{-n}\left(b\right)}\approx {\left(\frac{\mathrm{log}\left(1/b\right)}{\mathrm{log}\left(1/a\right)}\right)}^{n+1}.$
Let's prove this with an exact error term by making the first step precise. Let ${f}_{a}\left(x\right)={x}^{n}{a}^{x}$/ Then since
$\frac{{d}^{k}}{d{x}^{k}}{f}_{a}\left(x\right){|}_{x=0}=\frac{{d}^{k}}{d{x}^{k}}{f}_{a}\left(x\right){|}_{x=\mathrm{\infty }}=0$
for all k<n, Euler-Maclaurin summation up to $p=\frac{n}{2}$ for even n implies that
$|\sum _{k=0}^{\mathrm{\infty }}{k}^{n}{a}^{k}-{\int }_{0}^{\mathrm{\infty }}{x}^{n}{a}^{x}dx|\le |R|\le \frac{2\zeta \left(n\right)}{\left(2\pi {\right)}^{n}}\left({\int }_{0}^{\mathrm{\infty }}|{f}^{\left(n\right)}\left(x\right)|dx\right)\le \frac{2\zeta \left(n\right)}{\left(2\pi {\right)}^{n}}\left({2}^{n}\frac{\mathrm{\Gamma }\left(n+1\right)}{\mathrm{log}\left(1/a\right)}\right)=O\left(\frac{1}{{\pi }^{n}}\frac{\mathrm{\Gamma }\left(n+1\right)}{\mathrm{log}\left(1/a\right)}\right).$
Thus,
$\frac{{\text{Li}}_{-n}\left(a\right)}{{\text{Li}}_{-n}\left(b\right)}=\frac{\frac{\mathrm{\Gamma }\left(n+1\right)}{{\left(\mathrm{log}\left(\frac{1}{a}\right)\right)}^{n+1}}+O\left(\frac{1}{{\pi }^{n}}\frac{\mathrm{\Gamma }\left(n+1\right)}{\mathrm{log}\left(1/a\right)}\right)}{\frac{\mathrm{\Gamma }\left(n+1\right)}{{\left(\mathrm{log}\left(\frac{1}{b}\right)\right)}^{n+1}}+O\left(\frac{1}{{\pi }^{n}}\frac{\mathrm{\Gamma }\left(n+1\right)}{\mathrm{log}\left(1/b\right)}\right)}={\left(\frac{\mathrm{log}\left(1/b\right)}{\mathrm{log}\left(1/a\right)}\right)}^{n+1}\left(\frac{1+O\left({\pi }^{-n}\mathrm{log}\left(1/a{\right)}^{n}\right)}{1+O\left({\pi }^{-n}\mathrm{log}\left(1/b{\right)}^{n}\right)}\right),$
and so
$\frac{{\text{Li}}_{-n}\left(a\right)}{{\text{Li}}_{-n}\left(b\right)}={\left(\frac{\mathrm{log}\left(1/b\right)}{\mathrm{log}\left(1/a\right)}\right)}^{n+1}\left(1+O\left({\pi }^{-n}\mathrm{log}{\left(\frac{1}{min\left(a,b\right)}\right)}^{n}\right)\right),$
and using the fact that l$\mathrm{log}\left(1+x\right)=x+O\left({x}^{2}\right)$ , this implies implies that
$\mathrm{log}\left(\frac{{\text{Li}}_{-n}\left(a\right)}{{\text{Li}}_{-n}\left(b\right)}\right)=\left(n+1\right)\mathrm{log}\left(\frac{\mathrm{log}\left(1/b\right)}{\mathrm{log}\left(1/a\right)}\right)+O\left({\pi }^{-n}\mathrm{log}{\left(\frac{1}{min\left(a,b\right)}\right)}^{n}\right).$
Now, this error term is nontrivial only when
$1>a,b>{e}^{-\pi },$
and in this case it is $o\left({n}^{-N}\right)$ for any fixed capital N.

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