For which value of $x \in R $ the following series converges $ \sum_{n=1}^\infty n!x^n$

Question

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?

Answer 1

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.)

Answer 2

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