TITLE: How does this series scale ? (A fractional Touchard polynomial)
QUESTION [8 upvotes]: I was wondering how does series
$$
\sum_{k=1}^{\infty} \frac{x^k}{k!} \sqrt{k}
$$
scale with $x \rightarrow \infty$. And I am trying to prove that whether the limit
$$
\lim_{x \rightarrow \infty} \frac{\sum_{k=1}^{\infty} \frac{x^k}{k!} \sqrt{k}}{xe^x} = 0
$$
holds.

REPLY [6 votes]: Here is a bit different approach. Let $N_t$ be a Poisson process of the unit rate. Then
$$ \sum_{k=1}^{\infty} \frac{t^k}{k!} \sqrt{k} e^{-t} = \mathbb{E}[N_t^{1/2}]. $$
By the Jensen's inequality, we have
$$\mathbb{E}[N_t^{1/2}] \leq \mathbb{E}[N_t]^{1/2} = t^{1/2}.$$
To obtain the reverse inequality, observe that
$$ \mathbb{E}[N_t^{1/2}]
= t \sum_{k=0}^{\infty} \frac{t^k}{k!} \frac{1}{\sqrt{k+1}} e^{-t}
= t \mathbb{E}\left[ \frac{1}{\sqrt{N_t + 1}} \right] \geq \frac{t}{\sqrt{\mathbb{E}[N_t] + 1}} = \frac{t}{\sqrt{t+1}}. $$
where the last inequality again follows by Jensen's inequality.

Now let us investigate the asymptotic expansion. Write
\begin{align*}
N_t^{1/2}
&= t^{1/2}\left(1 + \frac{N_t - t}{t} \right)^{1/2} \\
&= t^{1/2} \sum_{k=0}^{N-1} \binom{1/2}{k} \left( \frac{N_t - t}{t}\right)^k
 + \mathcal{O}\left( \frac{1}{t^{N-1/2}} (N_t - t)^N \right). 
\end{align*}
Of course this only makes sense when $|N_t - t|$ is within fixed window of scale $t$, but the typical size of $N_t - t$ is of order $t^{1/2}$ and the Poisson distribution enjoys a very strong subgaussian tail. Thus modulo some probability theory stuffs, we get
\begin{align*}
\mathbb{E}[N_t^{1/2}]
&= t^{1/2} \sum_{k=0}^{N-1} \binom{1/2}{k} \frac{1}{t^k} \mathbb{E}\left[ (N_t - t)^k \right]
 + \mathcal{O}\left( \frac{1}{t^{\lceil N/2 \rceil- 1/2}} \right).
\end{align*}
For instance, if we plug $N = 3$ then we get the asymptotic expansion
$$ \sum_{k=1}^{\infty} \frac{t^k}{k!} \sqrt{k}
= \color{blue}{t^{1/2} e^t \left( 1 - \frac{1}{8t} + \mathcal{O}\left(\frac{1}{t^2}\right) \right)} \qquad \text{as} \quad t \to \infty.$$