For which values $x,\alpha$ does $\sum_{n=1}^{\infty}\frac{n^{nx}}{(n!)^ \alpha}$ converge? I want to know for which values of $x$ this series converges:
$$\sum_{n=1}^\infty \frac{n^{nx}}{(n!)^ \alpha};$$ here $\alpha \in\mathbb R$ is a constant
This series is defined $ \forall x \in \mathbb R$.
$$a_n=\frac{n^{nx}}{(n!)^ \alpha} \sim \frac{n^{nx}}{(\sqrt {2 \pi n}* (\frac{n}{e})^n)^ \alpha}= \frac{1}{(2 \pi n)^{\frac{\alpha}{2}}}n^{nx-n \alpha}e^{n \alpha} \sim 0 \Leftrightarrow  \alpha <0 \land x- \alpha <0.$$
Applying the root test:
$$ \sqrt[n]{\frac{1}{(2 \pi n)^{\frac{\alpha}{2}}}*n^{nx-n \alpha}*e^{n \alpha} }=\frac{1}{(2 \pi n)^{\frac{\alpha}{2n}}}n^{x- \alpha}e^{\alpha } <1 \iff x- \alpha<0   \iff x<\alpha $$
Is it right?
My doubt is regarding the necessary condition for the convergence in which I find
$\alpha <0 \land x- \alpha <0$ and the generality of $\alpha$ that doesn't make me say if $\alpha<$ or $>0$.

For these kind of sums,
I use $n! \sim (n/e)^n$ so
$\sum_{n=1}^\infty \frac{n^{nx}}{(n!)^ a}
\sim \sum_{n=1}^\infty \frac{n^{nx}}{(n/e)^{an}}\\
=\sum_{n=1}^\infty (\frac{e^an^{x}}{n^a})^n\\
=\sum_{n=1}^\infty (e^an^{x-a})^n\\
$
By the n-th root test, we want
$|e^an^{x-a}| < 1$
and this requires
$x < a$.
