<s>[INST] For which value of x $\sum_{n=1}^\infty \frac{n^{nx}}{n!}$ converges I want to know for which values of $x$ this series 
$$\sum_{n=1}^\infty \frac{n^{nx}}{n!}$$ converges.
This series is defined $ \forall x \in R$.
$a_n=\frac{n^{nx}}{n!}= \frac{(n^x)^{n}}{n!}= \frac{(n^x)}{n} \frac{(n^x)}{n-1} 
 \frac{(n^x)}{n-2}...\frac{(n^x)}{3}  \frac{(n^x)}{2}   \frac{(n^x)}{1}= n^{x-1} \frac{n^{x-1}}{1-\frac{1}{n}}...\frac{(n^x)}{3}  \frac{(n^x)}{2}   \frac{(n^x)}{1}
 \sim 0 \Leftrightarrow   x<1$.
Applying the ratio test:
$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{(n+1)x}}{(n+1)!} \frac{n!}{(n)^{nx}}=\frac{(n+1)^{(n+1)x}}{n+1} \frac{1}{(n)^{nx}}=\frac{(n+1)^{nx+x-1}}{n^{nx}} \sim \frac{(n)^{nx+x-1}}{n^{nx}}= (n)^{x-1}<1 \Leftrightarrow x-1<0   \Leftrightarrow x<1 $
Is it right?
 [/INST] For these I always use
$n! \sim (n/e)^n$.
So
$\sum_{n=1}^\infty \frac{n^{nx}}{n!}
\sim \sum_{n=1}^\infty \frac{n^{nx}}{(n/e)^n}
= \sum_{n=1}^\infty (en^x/n)^n
= \sum_{n=1}^\infty (en^{x-1})^n
$
so we want
$en^{x-1} < 1
$
or,
taking logs,
$1+(x-1)\ln(n) < 0$
or
$(x-1)\ln(n) < -1
$.
This is false if
$x \ge 1$,
and is true
if $x < 1$.
To check,
write $x = 1-c$.
the terms are
$\begin{array}\\
\frac{n^{nx}}{n!}
&\sim \frac{n^{n(1-c)}}{(n/e)^n}
&= \frac{n^{n(1-c)}e^n}{n^n}
&= n^{-cn}e^n
&= (n^{-c}e)^n
\end{array}
$
which goes to zero
only if $c > 0$.
</s>