Asymptotic behaviour of $f(x) = \sum_{n=1}^\infty n^\varepsilon \frac{x^n}{n!}$ for $\varepsilon\in(0,1)$

Let $\{a_n\}_{n=0}^{\infty}$ and $\{b_n\}_{n=0}^{\infty}$ be sequences of real numbers such that:

• $b_n$ is eventually positive (i.e. $b_n>0$ for all $n>n_0$);

• Both $a(x)=\sum\limits_{n=0}^{\infty}a_n x^n$ and $b(x)=\sum\limits_{n=0}^{\infty}b_n x^n$ converge for all $x$.

Then $\lim\limits_{n\to\infty}a_n/b_n=\lambda$ implies $\lim\limits_{x\to+\infty}a(x)/b(x)=\lambda$.

For a proof, we can assume $\lambda=0$ (otherwise replace $a_n$ with $a_n-\lambda b_n$). Now let $\varepsilon>0$, and $|a_n/b_n|<\varepsilon$ when $n>N$. Increasing $N$ if needed, we can assume $b_n>0$ when $n>N$. Then $$|a(x)|\leqslant\left|\sum_{n=0}^{N}a_n x^n\right|+\varepsilon\left(b(x)-\sum_{n=0}^{N}b_n x^n\right),$$ i.e. $\limsup\limits_{x\to+\infty}|a(x)/b(x)|\leqslant\varepsilon$. As $\varepsilon>0$ is arbitrary, we're done.

For $a>0$ we have $\Gamma(a)=\lim\limits_{n\to\infty}n^a\mathrm{B}(n, a)$.

This can be shown by taking $\lim\limits_{n\to\infty}$ of $$n^a\mathrm{B}(n,a)=n^a\int_0^1 t^{a-1}(1-t)^{n-1}\,dt=\int_0^n x^{a-1}(1-x/n)^{n-1}\,dx.$$

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Now write $f(x)=x\sum_{n=0}^{\infty}(n+1)^{\varepsilon-1}x^n/n!=xa(x)$ with $$a_n=\frac{(n+1)^{\varepsilon-1}}{n!},\qquad b_n=\frac{\mathrm{B}(n+1,1-\varepsilon)}{n!\ \Gamma(1-\varepsilon)}.$$ We have $\lim\limits_{n\to\infty}a_n/b_n=1$ and then $\lim\limits_{x\to+\infty}f(x)/(xb(x))=1$. But \begin{align}xb(x) &=\frac{x}{\Gamma(1-\varepsilon)}\sum_{n=0}^{\infty}\frac{x^n}{n!}\int_0^1 t^n(1-t)^{-\varepsilon}\,dt \\&=\frac{x}{\Gamma(1-\varepsilon)}\int_0^1(1-t)^{-\varepsilon}e^{xt}\,dt \\&=\frac{x^{\varepsilon}e^x}{\Gamma(1-\varepsilon)}\int_0^x z^{-\varepsilon}e^{-z}\,dz \end{align} (after substituting $t=1-z/x$). Thus $\color{blue}{\lim\limits_{x\to+\infty}f(x)/(x^{\varepsilon}e^x)=1}$.

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Yet another approach, allowing better asymptotics: \begin{align} f(x)&=x\sum_{n=0}^{\infty}\frac{x^n}{n!}(n+1)^{\varepsilon-1} \\&=\frac{x}{\Gamma(1-\varepsilon)}\sum_{n=0}^{\infty}\frac{x^n}{n!}\int_0^1 t^n(-\ln t)^{-\varepsilon}dt \\&=\frac{x}{\Gamma(1-\varepsilon)}\int_0^1 e^{xt}(-\ln t)^{-\varepsilon}dt \\&=\frac{xe^x}{\Gamma(1-\varepsilon)}\int_0^1 e^{-xt}\big(-\ln(1-t)\big)^{-\varepsilon}dt \end{align} and now we expand $\big(-\ln(1-t)/t\big)^{-\varepsilon}$ into power series: $$\big(-\ln(1-t)/t\big)^{-\varepsilon}=1-\frac{\varepsilon}{2}t-\frac{5\varepsilon-3\varepsilon^2}{24}t^2-\frac{6\varepsilon-5\varepsilon^2+\varepsilon^3}{48}t^3\pm\ldots$$ which, by Watson's lemma, gives $$f(x)\asymp x^{\varepsilon}e^x\left(1-\frac{\varepsilon(1-\varepsilon)}{2x}-\frac{\varepsilon(1-\varepsilon)(2-\varepsilon)(5-3\varepsilon)}{24x^2}-\frac{\varepsilon(1-\varepsilon)(2-\varepsilon)^2(3-\varepsilon)^2}{48x^3}\pm\ldots\right)$$

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.