Limits to infinity of a factorial function: $\lim_{n\to\infty}\frac{n!}{n^{n/2}}$

Question

How can this limit to infinity be solved? I've tried with d'Alembert but it just keeps coming up with the wrong answer.

$$\lim\limits_{n\to\infty}\frac{n!}{n^{n/2}}$$

I might have a problem in simplifying factorial numbers. Thank you in advance.

Answer 1

Hint: use Stirling's approximation: $$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}.$$

Answer 2

Using Stirling's Approximation

Stirling's Asymptotic Approximation says that $$ n!\sim\sqrt{2\pi n}\,\frac{n^n}{e^n}\tag{1} $$ This means that the expression in the question is $$ \frac{n!}{n^{n/2}}\sim\sqrt{2\pi n}\,\,\frac{n^{n/2}}{e^n}\tag{2} $$ which grows without bound. Therefore, $$ \lim_{n\to\infty}\frac{n!}{n^{n/2}}=\infty\tag{3} $$

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Another Approach

Squaring and writing the factorial forward and backward, for $n\ge4$, we get $$ \begin{align} \left(\frac{n!}{n^{n/2}}\right)^2 &=\overbrace{\frac{1(n-0)}{n}\frac{2(n-1)}{n}}^{\ge1}\overbrace{\frac{3(n-2)}{n}\cdots\frac{(n-2)3}{n}}^{\ge\left(\frac32\right)^{n-4}}\overbrace{\frac{(n-1)2}{n}\frac{(n-0)1}{n}}^{\ge1}\\ &\ge\left(\frac32\right)^{n-4} \end{align} $$ Each product in the numerator under the middle brace is $k(n-k+1)$. Since $k+(n-k+1)=n+1$ one of the numbers must be $\gt n/2$ while both are greater than $3$. Therefore, under the middle brace, $\frac{k(n-k+1)}n\gt\frac32$. For $n\ge2$, each of the terms under the outer braces is $\ge1$.

Answer 3

First note that for $n\geq 1$, we have $$n!\geq e\left(\frac{n}{e}\right)^n$$ $$\frac{n!}{n^{n/2}}\geq e\left(\frac{\sqrt n}{e}\right)^n$$ $$\lim\limits_{n\to\infty}\frac{n!}{n^{n/2}}\geq \lim\limits_{n\to\infty}e\left(\frac{\sqrt n}{e}\right)^n=\infty$$ Therefore $$\lim\limits_{n\to\infty}\frac{n!}{n^{n/2}}=\infty$$

Answer 4

$$n!\geq\left (\frac{n}{4}\right )^{3n/4}\\\lim_{n\to\infty}\frac{\left (\frac{n}{4}\right )^{3n/4}}{n^{n/2}}=\lim_{n\to\infty}\frac{n^{n/4}}{4^{3n/4}}=\lim_{n\to \infty}\left(\frac{n}{64}\right)^{n/4}=\infty$$ The inequality comes since $4n!=1\cdots \underbrace{n\cdots 4n}_{3n}\geq \underbrace{n\cdots n}_{3n}=n^{3n}$

Answer 5

HINT:

Without appealing to Stirling's Formula, we can write

$$\begin{align} \frac{(2n)!}{(2n)^{n}}&=\frac{2n(2n-1)(2n-2)\cdots (2n-(n-1))\cdot n(n-1)\cdots 3\cdot2\cdot1}{\underbrace{(2n)\cdots (2n)}_{n\,\,\text{copies}}}\\\\ &=\left(1/2+\frac{1}{2n}\right)\left(1/2+\frac{2}{2n}\right)\cdots \left(1/2+\frac{n-1}{2n}\right)\,n!\\\\ &\ge \frac{n!}{2^n} \end{align}$$

So, the problem boils down to showing that $\lim_{n\to \infty}\frac{n!}{2^n}=\infty$.

Answer 6

Consider:

$$n! = \int_0^\infty t^ne^{-t}\, dt > \int_n^{2n}t^ne^{-t}\, dt > n\cdot n^n e^{-2n} \ge n^ne^{-2n}.$$

Dividing by $n^{n/2}$ gives

$$ n!/n^{n/2} > n^{n/2}e^{-2n} = \exp [(n/2)\ln n - 2n].$$

Since $(n/2)\ln n - 2n \to \infty,$ the desired limit is $\infty.$