TITLE: Asymptotic behaviour of $f(x) = \sum_{n=1}^\infty n^\varepsilon \frac{x^n}{n!}$ for $\varepsilon\in(0,1)$
QUESTION [10 upvotes]: Let $\varepsilon \in (0, 1)$ and consider the analytic function $$f(x) = \sum_{n=1}^\infty n^\varepsilon \frac{x^n}{n!}.$$ 
What is the order of growth of $f(x)$ as $x \to \infty$?  From the basic inequality $1 \leqslant n^\varepsilon \leqslant n$ I deduced 
$$e^x - 1 < f(x) < xe^x$$ 
but I would like an asymptotic equivalence 
$$f(x) \sim g(x) e^x \qquad (x \to \infty)$$
where $g(x)$ is sufficiently familiar (e.g. a combination of $\log$s). How do I proceed?

REPLY [2 votes]: This is a sort of limit theorem for the Poisson distribution.
Note that $f(\lambda) =\sum_{n=1}^\infty n^\varepsilon \frac{\lambda^n}{n!} =\mathbb{E}[g(X)]e^{\lambda}$ for $g(x):=x^{\epsilon}$ and a Poisson random variable $X$ with mean $\lambda$.
Lemma: $$|\mathbb{E}[g(X)]-g(\mathbb{E}[X])|\leq |\epsilon(\epsilon-1)|/2\lambda^{\epsilon-1}$$
Proof: $|g(x)-g(\lambda)-g'(\lambda)(x-\lambda)|\leq  |g''(\lambda)/2|(x-\lambda)^2=|\epsilon(\epsilon-1)|\lambda^{\epsilon-2}/2(x-\lambda)^2$. Hence
$$
|\mathbb{E}[g(X)]-g(\mathbb{E}[x])|=|\mathbb{E}[g(X)-g(\lambda)-g'(\lambda)(X-\lambda)]|\leq |\epsilon(\epsilon-1)|\lambda^{\epsilon-2}/2\mathbb{V}[X]\leq |\epsilon(\epsilon-1)|\lambda^{\epsilon-1}/2
$$
where we used that the variance of the Poisson distribution is $\lambda$.
Conclusion: $$f(\lambda)/(\lambda^{\epsilon}e^{\lambda})\to 1$$
Note that this works for all $\epsilon\in\mathbb{R}$. 
You can also get more asymptotic terms by using more terms in the Taylor expansion of $g$ and using known formulas for the centered moments of the Poisson distribution.