Is this relationship, $\sum^{\infty}_{N=1}\frac{N^{-\alpha} \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^{-\alpha}$, true? According to numerical simulation, the relationship
$$\sum^{\infty}_{N=1}\frac{N^{-\alpha} \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^{-\alpha}$$
where $\Gamma$ is the Gamma function seems to be true.
Do you have any idea how to show this relationship using asymptotic or exact methods?
 A: Put $\mu:=\alpha-1$; then $\sum^{\infty}_{n=1}\frac{ x^{n-1}}{(n-1)!n^\alpha} = x^{-1}\sum^{\infty}_{n=1}\frac{ x^n}{n!n^\mu}=x^{-3/2}I_\mu(x)$ for the function $I_\mu(x)$ given in Johannes Trost’s answer to Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$, with a complete asymptotic series.
A: Let $b:=-\alpha\in\mathbb C$. We have to show that
$$f(x):=\sum^{\infty}_{N=1}\frac{N^b \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^b \tag{1}$$
as $x\to\infty$.
Let $Y=Y_x$ denote a random variable with the Poisson distribution with parameter $x$. Using the idea of this answer, we have
$$f(x)=e^x x^b\, E\Big(\frac{Y+1}x\Big)^b. \tag{2}$$
Next, $\dfrac{Y+1}x\to1$ in probability (as $x\to\infty$), and $EY(Y-1)\cdots(Y-(k-1))=x^k$ and hence $EY^k\sim x^k$ for all natural $k$. So, by the uniform integrability of $\big(\frac{Y+1}x\big)^b$ (see e.g. the de la Vallée-Poussin theorem),
$$E\Big(\frac{Y+1}x\Big)^b\to1.$$
Thus, (1) follows from (2).

The uniform integrability of $\big(\frac{Y+1}x\big)^b$ also follows because
$$E\exp\frac{Y+1}x=\exp\Big\{\frac1x+x(e^{1/x}-1)\Big\}\to e<\infty.$$
