Prove that $\sum_{n=1}^{\infty}n!x^n$ only converges if $x=0$

Question

The question title basically sums up the goal of this question, that is, how to show that the power series $$\sum_{n=0}^{\infty}n!x^n $$ converges only if $x=0$.

I will show my closest attempt at proving this, using the root test as all other methods, like divergence test or ratio test, seemed to be inconclusive or out of reach for me to prove.

Attempt

Although I cannot prove this (and it would be amazing if someone could offer a proof outline or show a proof of this), it is known that $$n!\geq \left(\frac{n}{3}\right)^n\tag{1}$$ for all $n\in\mathbb{N}$ which gives $$\sqrt[n]{n!}\geq \frac{n}{3}\tag{2}$$ for all $n\in\mathbb{N}$.

Assuming this, let $x\neq 0$. Then $$\lim_{n\to\infty}\sqrt[n]{n!|x|^n }=|x|\cdot\lim_{n\to\infty}\sqrt[n]{n!}=\infty $$ since $$\lim_{n\to\infty}\frac{n}{3}=\infty$$ thus the power series diverges by the root test.

Questions

I have two questions.

First: Is this valid assuming $(1)$ and $(2)$?

Second: How would one prove $(1)$ and $(2)$. I cannot do it inductively. Also, if there is a different or simpler method, please let me know or guide me in the right direction. Thanks

Answer 1

Yes, the proof is corect if (1) and (2) aer assumed. To prove (1) use induction. To go from $n$ to $n+1$ you will need $3 \geq (1+\frac 1 n)^{n}$. For this note that $(1+\frac 1 n)^{n}=e^{n \log (1+\frac 1 n)} \leq e^{1}<3$.

Answer 2

In this case, you don't really need to use anything other than the fact that a series can only converge if its general term goes to zero. When $|x| \ge 1$, it is immediate to conclude that $\displaystyle \lim_{n \to \infty} n! x^n \ne 0$ and, therefore the series is divergent. If we show that when $x \in (-1,1)\setminus \{0\}$ the limit is also not zero, we are done. Denoting $y = \frac 1x$, we want to analyse $$ \lim \dfrac{n!}{y^n}, \quad |y| > 1. $$

For $y>1$, using Stirling's formula, we get $$ \lim \dfrac{n!}{y^n} = \lim \dfrac{\sqrt{2 \pi n} (n/e)^n}{y^n} = \lim \sqrt{2 \pi n} \left(\frac{n \cdot e}{y} \right)^n = +\infty. $$