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Solving the problem on Quora, I tried also to get a reasonable approximation of $$ (S_n)^\frac1n=\left(\sum_{k=0}^\infty\frac {k^n}{k!}\right)^\frac1n\,\text{at}\,n\to\infty$$ Using Laplace method (which, frankly speaking, is not formally applicable in this case), I got $$(S_n)^\frac1n\sim \frac {n\,e^{\frac1{W(n)}}}{e\,W(n)}$$ where $W(n)$ is Lambert function. The approximation works rather well - in spite of arbitrary usage of the method: $\displaystyle n=100 \quad\frac{\ln n}n\left(S_n\right)^\frac1n\bigg|_{n=100}=0.667365...\,;\quad \frac {\ln n\,\,e^{\frac1{W(n)}}}{e\,W(n)}\bigg|_{n=100}=0.672337...$

$\displaystyle n=1000 \quad\frac{\ln n}n\left(S_n\right)^\frac1n\bigg|_{n=1000}=0.585123...\,;\quad \frac {\ln n\,\,e^{\frac1{W(n)}}}{e\,W(n)}\bigg|_{n=1000}=0.585692...$

etc.

My question is whether we can prove the result in a rigorous way, or provide a more accurate and justified asymptotic, using some other evaluation technics?


Follow up

The asymptotic of Bell numbers solves the problem; @Command Master and @Cactus - thank you for the information. One of the ways of evaluation (by means of complex integration) can be found in this pape. It is interesting to note that the correct first term of the asymptotics can be obtained in one line by simply applying the Laplace method. I got $$\sum_{k=0}^\infty\frac {k^n}{k!}=\frac1{\sqrt{W(n)+1}}\left(\frac {n\,e^{\frac1{W(n)}}}{e\,W(n)}\right)^n$$ what coincides with the first term of the asymptotics in the paper. The formula in Wikipedia is not quite accurate (contains $\frac1{\sqrt{W(n)}}$ instead of $\frac1{\sqrt{W(n)+1}}$).

The formula gives good approximation already for small $n\sim 10$.

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    $\begingroup$ You can use Dobinski's formula and the known asymptotic growth rate of the Bell numbers $\endgroup$ Sep 21, 2023 at 15:18

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First of all, you can simplify you equivalent. Indeed, $W(n) \rightarrow +\infty$ thus $e^{1/W(n)} \sim 1$ and $W(n) \sim \ln(n)$ thus $\frac{ne^{1/W(n)}}{eW(n)} \sim \frac{n}{e\ln(n)}$.

Then, as @Command Master suggested it, you can use the Dobinski formula, $$ B_n = \frac{1}{e}\sum_{k \geqslant 0} \frac{k^n}{k!}, $$ where the $(B_n)$ are the Bell numbers (see https://en.wikipedia.org/wiki/Bell_number), which verify, $$ B_n \sim \frac{1}{\sqrt{n}}\left(\frac{n}{W(n)}\right)^{n + 1/2}\exp\left(\frac{n}{W(n)} - n - 1\right). $$ Therefore, $$ S_n^{1/n} = (eB_n)^{1/n} \sim \frac{1}{n^{1/(2n)}}\left(\frac{n}{W(n)}\right)^{1 + 1/(2n)}\exp\left(\frac{1}{W(n)} - 1\right) \sim \frac{n}{e\ln(n)}. $$ Notice that $u_n \sim v_n > 0$ for $n$ large enough implies $u_n^{1/n} \sim v_n^{1/n}$ but the reciprocal is false in general, so having an equivalent on $S_n$ (given by the equivalent of Bell's numbers) is much more powerful than having an equivalent on $S_n^{1/n}$.

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    $\begingroup$ Right, I corrected it. I don't know how we get the asymptotic, it seems pretty hard, I'll try to find an other way to prove the equivalent of $S_n^{1/n}$ (which is easier). $\endgroup$
    – Cactus
    Sep 21, 2023 at 17:40

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