Consider the power series $\sum_{k=0}^\infty \sqrt k \frac{x^k}{k!}$. It is easily seen that its radius of convergence is $\infty$. I'm looking for an elementary proof of the asymptotic expansion $$\sum_{k=0}^\infty \sqrt k \frac{x^k}{k!}=e^x\left(\sqrt x - \frac{1}{8 \sqrt x} + o\left(\frac{1}{\sqrt x} \right)\right)$$ as $x$ goes to $\infty$.

Using Stirling's estimate and an asymptotic property of power series, one may derive $$\sum_{k=0}^\infty \sqrt k \frac{x^k}{k!}\sim \frac{1}{\sqrt{ 2\pi}}\sum_{k=0}^\infty \frac{(ex)^k}{k^k}$$

so the question boils down to finding an estimate of $$\sum_{k=0}^\infty \frac{x^k}{k^k}$$

I find this answer quite unconvincing since it makes heavy use of asymptotics of special functions and is not very rigorous .

I'd be satisfied if someone showed how to derive the simpler estimate $$\sum_{k=0}^\infty \sqrt k \frac{x^k}{k!}\sim e^x \sqrt x$$


1 Answer 1


This answer deals with different approaches to the asymptotics of the series in question. Here I will try to give a probabilistic argument.

This series has the following probabilistic interpretation: If $N_t$ is a Poisson distribution of rate $t > 0$ then we have

$$ \mathbb{E}[\sqrt{N_t}] = \sum_{k=0}^{\infty} \sqrt{k} \, \frac{t^k}{k!}e^{-t}. $$

Together with some basic inequalities we can get the leading order of the asymptotics. For instance, applying Jensen's inequality with the concave function $x \mapsto \sqrt{x}$ tells that

$$ \mathbb{E}[\sqrt{N_t}] \leq \sqrt{\mathbb{E}[N_t]} = \sqrt{t} $$

whereas writing $ \mathbb{E}[\sqrt{N_t}] = t \sum_{k=0}^{\infty} \frac{t^k}{k!} \frac{1}{\sqrt{k+1}} e^{-t} = t \mathbb{E}\left[ (N_t + 1)^{-1/2} \right] $ and applying Jensen's inequality with the convex function $x \mapsto (x+1)^{-1/2}$ gives

$$ \mathbb{E}[\sqrt{N_t}] \geq \frac{t}{\sqrt{\mathbb{E}[N_t] + 1}} = \frac{t}{\sqrt{t+1}}. $$

In terms of the original series, this reads as

$$ \frac{x}{\sqrt{x+1}} e^x \leq \sum_{k=0}^{\infty} \sqrt{k} \, \frac{x^k}{k!} \leq \sqrt{x} e^x, $$

which is enough to derive the asymptotics $\sim \sqrt{x} e^x$. Higher order terms can also be extracted by utilizing various concentration behaviors of $N_t$.

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    $\begingroup$ So this is a duplicate ! Thanks for providing the link to the original question, as well as a new approach. $\endgroup$ Nov 2, 2017 at 9:43
  • $\begingroup$ @GabrielRomon Thank you for accepting this! Also I would consider it a border case, and personally I think it is a good chance to invite other users to ponder about it and provide self-contained and not-hard-to-swallow arguments. All the answers in the link (including my answer there) are either not self-contained or kind of hand-waving, which are probably not what you wanted. $\endgroup$ Nov 2, 2017 at 9:48

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