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?



Is this relationship, $\sum^{\infty}_{N=1}\frac{N^{-\alpha} \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^{-\alpha}$, true?
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 $\sum\limits\_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$](https://mathoverflow.net/questions/264595/asymptotic-expansion-of-sum-limits-n-1-infty-fracx2n1n-sqrtn/266265#266265), with a complete asymptotic series.


