Find $\lim_{n \to \infty} \sqrt[n]{n!}$.

Question

I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered $\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}$. I have made the claim that $\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.$ I have begun the proof and I think it is on the right track but some verification would be nice.

Proof: We can see that for n=1, this obviously holds, $\left(1!\right)^1\geq 1.$ Now suppose that this is the case for some $n=k$, then we have $\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}$. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks!

Answer 1

I'm not sure it holds in general. Consider $n! = n(n-1)(n-2)...2\cdot1$, at least $n/2$ terms are $>n/2$ thus; $$n! > \left(\frac{n}{2}\right)^\frac{n}{2}$$ taking the reciprocal $$ \frac{1}{n!} < \left(\frac{2}{n}\right)^\frac{n}{2}$$ then raising to the power $1/n$ gives $$ \left(\frac{1}{n!}\right)^\frac{1}{n} < \left(\frac{2}{n}\right)^\frac{1}{2}$$ clearly the right hand side has $\lim_{n\rightarrow \infty} \sqrt{2/n} = 0$ thus $$\lim_{n\rightarrow \infty} \left(\frac{1}{n!}\right)^\frac{1}{n} = 0$$

Edit: of course njguliyev's comment is a much easier way to see it can't be $\geq1$.

Answer 2

$$ \lim_{n \to \infty}{-\ln\left(\Gamma\left(n + 1\right)\right) \over n} = -\lim_{n \to \infty}\Psi\left(n + 1\right) = -\infty $$

$$ \color{#ff0000}{\lim_{n \to \infty}\left(1 \over n!\right)^{1/n} = 0} $$

$\Gamma$ and $\Psi$ are the Gamma and Digamma functions, respectively.

$\mbox{Or Stirling dixit:}$ \begin{align} \lim_{n \to \infty}\left(1 \over n!\right)^{1/n} & = \lim_{n \to \infty}\left(1 \over \sqrt{2\pi\,}\,n^{n + 1/2}\, {\rm e}^{-n}\right)^{1/n} = {1 \over \sqrt{2\pi\,}\,}\, \lim_{n \to \infty}{{\rm e} \over n^{1 + 1/2n}} \\[5mm] & = {{\rm e} \over \sqrt{2\pi\,}\,}\, \lim_{n \to \infty}\left(1 \over n\right) = \color{#ff0000}{0} \end{align}

Answer 3

The question seems to be about $n!^{1/n}\ge 1$. This is fairly obvious, as $n!\ge1$. There is no need for induction.

Answer 4

For an elementary approach which gives more information, you can start with $(1+1/n)^n \lt e \lt (1+1/n)^{n+1} $. This is proved by showing that $(1+1/n)^n$ is an increasing function of $n$ and $(1+1/n)^{n+1}$ is an decreasing function of $n$ and that they have a common limit, good old $e$.

Using this and induction, you can show that $(n/e)^n \lt n! < (n/e)^{n+1} $ so that $n/e \lt (n!)^{1/n} < (n/e)(n/e)^{1/n} $.

Since both $n^{1/n}$ and $a^{1/n}$ approach $1$ as $n \to \infty$, $\dfrac{(n!)^{1/n}}{n} \to \dfrac1{e} $.

Here are the details.

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Here's the proof that $(1+1/n)^n \lt e \lt (1+1/n)^{n+1} $.

Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1} $. We will prove that $a_n$ is an increasing sequence and $b_n$ is an decreasing sequence. Since $a_n < b_n$, this implies, for any positive integers $n$ and $m$ with $m < n$ that $a_m < a_n < b_n < b_m$.

We use the very ingenious proof in [1], which uses the arithmetic-geometric mean inequality (AGMI), which we will use in the form $((v_1+v_2+...v_n)/n)^n \gt v_1v_2...v_n $ (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.

For $a_n$, consider $n$ values of $1+1/n$ and $1$ value of $1$. By the AGMI, $((n+2)/(n+1))^{n+1} > (1+1/n)^n $, or $(1+1/(n+1))^{n+1} > (1+1/n)^n $, or $a_{n+1} \gt a_n $.

For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$. By the AGMI, $(n/(n+1))^{n+1} \gt (1-1/n)^n$ or $(1+1/n)^{n+1} < (1+1/(n-1))^n $, or $b_n < b_{n-1} $ so $b_{n+1} < b_{n} $.

Since $b_n-a_n =(1+1/n)^{n+1}-(1+1/n)^n =(1+1/n)^{n}(1+1/n-1) =(1+1/n)^{n}/n \to 0 $, $a_n$ and $b_n$ have a common limit, which I will call $e$.

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Here's the proof that $(1+1/n)^n \lt e \lt (1+1/n)^{n+1} $ implies that $(n/e)^n \lt n! \lt (n/e)^{n+1} $.

Suppose $n! > (\dfrac{n}{e})^n $. We want to show that $(n+1)! > (\dfrac{n+1}{e})^{n+1} $.

Then

$\begin{array}\\ (n+1)! &=(n+1)n!\\ &>(n+1) (\dfrac{n}{e})^n\\ &=\dfrac{(n+1)n^n}{e^n}\\ \end{array} $

so we want $\dfrac{(n+1)n^n}{e^n} \gt \dfrac{(n+1)^{n+1}}{e^{n+1}} $ or $e \gt \dfrac{(n+1)^n}{n^n} =(1+1/n)^n $ which is shown above.

Similarly, suppose $n! < (\dfrac{n}{e})^{n+1} $. We want to show that $(n+1)! < (\dfrac{n+1}{e})^{n+2} $.

Then

$\begin{array}\\ (n+1)! &=(n+1)n!\\ &<(n+1) (\dfrac{n}{e})^{n+1}\\ &=\dfrac{(n+1)n^{n+1}}{e^{n+1}}\\ \end{array} $

so we want $\dfrac{(n+1)n^{n+1}}{e^{n+1}} \lt \dfrac{(n+1)^{n+2}}{e^{n+2}} $ or $e \lt \dfrac{(n+1)^{n+1}}{n^{n+1}} =(1+1/n)^{n+1} $ which is shown above.

[1] N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.

Answer 5

We have that

$$\sqrt[n]{n!}=e^{\frac{\log (n!)}n} \to \infty$$

indeed by Stolz-Cesaro theorem

$$\frac{\log ((n+1)!)-\log(n!)}{n+1-n}=\log (n+1) \to \infty \implies \frac{\log (n!)}n \to \infty$$