$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

Question

How do I prove that $ \displaystyle\lim_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite?

Answer 1

By considering Taylor series, $\displaystyle e^x \geq \frac{x^n}{n!}$ for all $x\geq 0,$ and $n\in \mathbb{N}.$ In particular, for $x=n$ this yields $$ n! \geq \left( \frac{n}{e} \right)^n .$$

Thus $$\sqrt[n]{n!} \geq \frac{n}{e} \to \infty.$$

Answer 2

$n! \geq (n/2)^{n/2}$ because half of the factors are at least $n/2$. Take $n$-th root.

Answer 3

Using $\text{AM} \ge \text{GM}$

$$ \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n} \ge \sqrt[n]{\frac{1}{n!}}$$

$$\sqrt[n]{n!} \ge \frac{n}{H_n}$$

where $H_n = 1 + \frac{1}{2} + \dots + \frac{1}{n} \le \log n+1$

Thus $$\sqrt[n]{n!} \ge \frac{n}{\log n+1}$$

Answer 4

(EDIT: first lines reworded due to criticism below)

We wish to show that for a fixed real number $\alpha$, we have $$ (n!)^{\frac{1}{n}}>\alpha $$ for sufficiently large $n$. Clearly it suffices to show that the logarithm of this quantity exceeds $\alpha$ (for sufficiently large $n$).

Using elementary log rules, we have \begin{align} \log(n!^{\frac{1}{n}}) &= \frac{\log(n!)}{n}\\ &= \frac{1}{n}\sum_{i=1}^{n} \log(i)\\ \end{align} We will start the sum at $2$ since $\log(1)$ is 0. Now $$ \frac{1}{n}\sum_{i = 2}^{n} \log(i) $$ is a right-handed Riemann sum (with step-size $\Delta x$ = 1) which gives an overestimate for the integral $$ \int_1^n\log(x)dx. $$ But this integral is just $$ n\log(n) - n + 1. $$ Thus we have shown that $$ \log(n!^{\frac{1}{n}}) \geq \log(n) - 1 + \frac{1}{n} $$ and the right hand side goes to $\infty$ as $n$ goes to $\infty$, which is what we want.

Answer 5

Using a multiplicative variant of Gauss's trick we get: $$ (n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n $$ Hence $$ \sqrt[n]{n!} \ge \sqrt{n} \to \infty $$

Answer 6

Hint :

$$n! \approx \sqrt{2\pi n}\cdot\left(\frac{n}{e}\right)^n$$

Answer 7

Denote $S_n=\sqrt[n]{n!}$. Then $$ \log S_n= \frac{\log 1+..+\log n}{n}=\frac{a_n}{b_n}$$.

Since $b_n$ is increasing and tends to $\infty$ and we have $$ \frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\log(n+1) \to \infty$$ we can apply Stolz-Cesaro theorem and conclude that the limit $L=\lim_{n \to \infty} \log S_n$ exists and $L=\infty$. This implies that $S_n \to \infty$.

Answer 8

A completely elementary proof: try proving that $n! > 2^n$ for $n$ sufficiently large, then $n! > 3^n$ for $n$ even larger, and so on, so eventually you'll have for each $k$ there is some $n$ such that $n! > k^n$, i.e. $\sqrt[n]{n!} > k$, which is enough to prove what you asked.

Intuitively it's obvious that $n! > k^n$: as we go from $k^n$ to $k^{n+1}$ we multiply by $k$, but going from $n!$ to $(n+1)!$ multiplies by $n+1$ which will eventually be far larger than $k$. In fact, by the time you reach $n = k^2$, we're growing twice as fast as $k^n$, which ought to give us a crude bound for when we must pass it: $(k^2 + l)!$ is a product involving $l$ terms greater than $k^2$, so it is certainly bigger than $(k^2)^l = k^{2l}$, so setting $l = k^2$, $(2k^2)! > k^{2k^2}$.

Of course, if you actually check the numbers, the factorial beats the exponentiation way sooner than that, but that's irrelevant to the proof.

Answer 9

In light of sdcvvc' answer, this answer may be a bit much; but you can generalize the following argument to show that if $(a_n)$ is a sequence of positive numbers and if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=\infty$, then $\lim\limits_{n\rightarrow\infty}{\root n\of{a_n}}=\infty$. (More generally, one can show that $\limsup\limits_{n\rightarrow\infty}{\root n\of{a_n}} \le \limsup\limits_{n\rightarrow\infty}{a_{n+1}\over a_n} $ and that $ \liminf\limits_{n\rightarrow\infty}{a_{n+1}\over a_n} \le \liminf\limits_{n\rightarrow\infty}{\root n\of{a_n}} $. )

Let $a_n=n!$. One can show by induction that $a_{n+k}\ge n^k a_n$ for all positive integers $n$ and $k$.

Now fix a positive integer $N$ and let $n$ be a positive integer with $n\ge N$. Then $$\tag{1} a_n =a_{N+(n-N)} \ge N^{n-N} a_N=N^n\cdot {a_N\over N^N},\qquad\qquad(n\ge N). $$ Taking the $n^{\rm th}$ roots of the left and right hand sides of $(1)$ gives $$\tag{2} \root n\of{a_n}\ge N\cdot{\root n\of {a_N}\over (N^N)^{1/n}}, \qquad\qquad(n\ge N). $$ Now, as $n\rightarrow\infty$, the righthand side of $(2)$ tends to $N$. From this it follows that $\liminf\limits_{n\rightarrow\infty} \root n\of{a_n}\ge N$. But, as $N$ was arbitrary, we must then have $\lim\limits_{n\rightarrow\infty} \root n\of{a_n}=\infty$.

Answer 10

This solution is somehow similar to Beni's:

We prove a stronger statement: $\lim_n \frac{\sqrt[n]{n!}}{n}=\frac{1}{e}$.

Proof:

$$\lim_n \frac{\sqrt[n]{n!}}{n}=e^{ \lim_n \ln(\frac{\sqrt[n]{n!}}{n})} $$

We calculate now separately

$$\lim_n \ln(\frac{\sqrt[n]{n!}}{n})= \lim_n \frac{\ln \frac{n!}{n^n}}{n}$$

By Stolz Cezaro we get

$$\lim_n \ln(\frac{\sqrt[n]{n!}}{n})= \lim_n\frac{\ln \frac{(n+1)!}{(n+1)^{n+1}}-\ln \frac{n!}{n^n}}{n+1-n}= \lim_n\ln \frac{n!}{(n+1)^{n}}-\ln \frac{n!}{n^n}= \lim_n\ln \frac{n^n}{(n+1)^{n}}$$ $$=\lim_n\ln \frac{1}{(1+\frac1n)^{n}}=\ln(\frac{1}{e})=-1$$

Answer 11

Let's start out with a nice result expressed in harmonic numbers from one of the posted problems, see here. When n is very large we get that:

$$ (n+1) H_n - n \approx \ln n!\longrightarrow e^{(n+1) H_n - n} \approx n!\longrightarrow e^\frac{{(n+1) H_n - n}}{n} \approx \sqrt[n]{n!} \longrightarrow \infty$$

REMARK: $\lim_{n\to\infty} \frac{(n+1) H_n - n}{\ln n!}=1$ may be proved by Cesaro-Stolz theorem such that we may avoid the use of Stirling's approximation.

The proof is complete.