<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 $ n\in\mathbb{N}^{*} $, denoting $ f_{n}:x\mapsto\frac{n^{nx}}{n!} \cdot $
If $ x\geq 1 $, observe that $ f_{n}\left(x\right)\geq\frac{n^{n}}{n!}=\prod\limits_{k=1}^{n-1}{\frac{n}{k}}\geq\prod\limits_{k=1}^{n-1}{\frac{k+1}{k}}=n $, meaning $ \lim\limits_{n\to +\infty}{f_{n}\left(x\right)}=+\infty\neq 0 $, which means $ \sum\limits_{n\geq 1}{f_{n}\left(x\right)} $ diverges.
If $ x<1 $, using d'Alembert's ratio test, $ \lim\limits_{n\to +\infty}{\frac{f_{n+1}\left(x\right)}{f_{n}\left(x\right)}}=\lim\limits_{n\to +\infty}{\frac{\left(n+1\right)^{\left(n+1\right)x}}{n^{nx}\left(n+1\right)}}=\lim\limits_{n\to +\infty}{\left(n+1\right)^{x-1}\left(1+\frac{1}{n}\right)^{nx}}=0<1 $, meaning $ \sum\limits_{n\geq 1}{f_{n}\left(x\right)} $ converges.
</s>