For which value of $x \in R $ the following series converges $ \sum_{n=1}^\infty n!x^n$ For which value of $x \in R $ the following series converges
$ \sum_{n=1}^\infty n!x^n$
The series of the absolute values is $ \sum_{n=1}^\infty n! |x|^n$
and applying the root test:
$ lim_{n\rightarrow \infty} \sqrt[n]{ n! |x|^n}= lim_{n\rightarrow \infty}  (n!)^{ \frac{1}{n}} |x| =|x|<1 $
because I think that $ lim_{n\rightarrow \infty}  (n!)^{ \frac{1}{n}}  =1 $
$(n*(n-1)*(n-2)...*3 *2* 1)^{ \frac{1}{n}}=n^{ \frac{1}{n}}*(n-1)^{ \frac{1}{n}}*(n-2)^{ \frac{1}{n}}*...*3^{ \frac{1}{n}}*2^{ \frac{1}{n}}*1^{ \frac{1}{n}} 
 \rightarrow 1*1*1*...*1*1*1=1$
while the solution should be $x=0$
Can someone help me to understant where is the mistake?
 A: Here is your mistake:

because I think that $\lim_{n\rightarrow \infty}  (n!)^{ \frac{1}{n}}  =1$

Actually, we have 
$$\lim_{n\rightarrow \infty}  (n!)^{ \frac{1}{n}}  = +\infty\,.$$
This is not hard to see e.g. using Stirling's approximation (or even more elementary, cruder inequalities on $\log n!$ will do):
$$
\log\left( (n!)^{ \frac{1}{n}} \right) = \frac{1}{n} \log n!
= \frac{1}{n}\left(n\log n +O(n)\right)
= \log n +O(1) \xrightarrow[n\to\infty]{} \infty\,.
$$
(The above argument with Stirling's approximation actually yields $ (n!)^{ \frac{1}{n}} = \Theta(n)$, so it goes to $\infty$ at a linear rate.)
A: 
Can someone help me to understand where is the mistake?

You have applied a limit rule concerning finite products to an infinite product, which is no longer true.
One may recall that
$$
\left(1+\frac1k \right)^k<e,\qquad k\ge1,
$$ then by multiplying this over $k=1,2,\cdots,n-1$, factors telescope, one obtains
$$
\frac{n^n}{n!}<e^n
$$ or

$$
\frac{n}{e}<(n!)^{\large\frac1n}
$$ 

giving
$$
\lim_{n\to \infty}(n!)^{\large\frac1n}=\infty.
$$
