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 $>0$.
 A: Your approach is correct! Here's an other approach using the Ratio Test.
Let's first consider the case $x=\alpha$:
$$
\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)\alpha}}{n^{n\alpha}}\cdot\frac{(n!)^{\alpha}}{((n+1)!)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)\alpha}}{n^{n\alpha}}\cdot\frac{1}{(n+1)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{n\alpha}}{n^{n\alpha}}\right|
$$By continuity, we can pull the $\alpha$ powers outside the limit, which is then well-known:
$$
=\left(\lim_{n\to\infty}\left|\frac{(n+1)^{n}}{n^{n}}\right|\right)^{\alpha} =\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^{\alpha} =e^{\alpha}
$$So when $x=\alpha$, the series converges only when $\alpha<0$. The other two cases can be deduced the same way:
$$
\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)x}}{n^{nx}}\cdot\frac{(n!)^{\alpha}}{((n+1)!)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{nx}}{n^{nx}}\cdot\frac{(n!)^{\alpha}(n+1)^x}{((n+1)!)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{nx}}{n^{nx}}\cdot(n+1)^{x-\alpha}\right|
$$
$$
=e^x\lim_{n\to\infty}\left|(n+1)^{x-\alpha}\right|
$$Clearly this converges only when $x<\alpha$. So in summary, either $x<\alpha$, or $x=\alpha <0$.
A: 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
$x < a$.
