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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

C Squared
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  • Inequality $(1)$ should follow, at least asymptotically, from Stirling's formula... but maybe there is a more direct proof... – hgmath Feb 17 '21 at 23:17
  • @hgmath I was kind of hoping there was, since I don't know to much about asymptotics. – C Squared Feb 17 '21 at 23:18
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    While I trust you that the $n/3$ lower bound is accurate, I only know an easy proof for a lower bound of $(n/2)^n$. It goes like this: Pair up $(1,n)$, $(2, n-1)$, and so on - each product is at least $(n+1)^2/2$ by AM-GM. Multiply them all together. – preferred_anon Feb 17 '21 at 23:18
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    @preferred_anon I'll see if I can work with this too. thanks – C Squared Feb 17 '21 at 23:19
  • Actually, the ratio test works, since $\lim_{n\to\infty}|(n+1)x|=\infty>1$ for $x\ne0$. – J.G. Feb 17 '21 at 23:23
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    @J.G. wikipedia states that "if $L = 1$ or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case." Since the limit fails to exists, I don't think I can use it. The root test however, does account for the infinite case. – C Squared Feb 17 '21 at 23:25
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    @CSquared It being $\infty$ is fine, but maybe you'd be better noting the terms increase in modulus so don't tend to $0$. – J.G. Feb 18 '21 at 07:20
  • $$ e^n = 1 + n + \cdots + \frac{{n^n }}{{n!}} + \cdots > \frac{{n^n }}{{n!}} \Rightarrow n! > \left( {\frac{n}{e}} \right)^n \Rightarrow n! > \left( {\frac{n}{3}} \right)^n $$ – Gary Feb 18 '21 at 11:33
  • Related (just linking it) – metamorphy Feb 18 '21 at 14:06

2 Answers2

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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 $(1+\frac 1 n)^{n}=e^{n \log (1+\frac 1 n)} \leq e^{1}<3$.

Kavi Rama Murthy
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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 $$ \lim \dfrac{n!}{y^n} = \lim \dfrac{\sqrt{2 \pi n} (n/e)^n}{y^n} = \lim \sqrt{2 \pi n} \left(\frac{n \cdot e}{y} \right)^n = +\infty. $$

PierreCarre
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