5
$\begingroup$

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 orderded 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 exprerimentation 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?

$\endgroup$
4
$\begingroup$

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 $N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.