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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.

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

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    $\begingroup$ Is this the only way to solve this?Can you please provide some alternative way may be using inequalities etc. $\endgroup$ – Learnmore Nov 19 '16 at 18:04
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    $\begingroup$ I think this is the easiest way $\endgroup$ – E.H.E Nov 19 '16 at 18:12
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    $\begingroup$ It's easy if you know it. If you don't, it's not a way at all. $\endgroup$ – zhw. Nov 20 '16 at 1:20
  • $\begingroup$ Exactly,I second that @zhw.;I don't know this formula at all $\endgroup$ – Learnmore Nov 20 '16 at 2:54
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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 $\lim_{n\to \infty}\frac{n!}{2^n}=\infty$.

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  • $\begingroup$ Aren't there some problems? 1. In the Numerator will not it be $2n-(n-1)$?2.In the second step will not it be $\frac{1}{2}-\frac{1}{2n}$? $\endgroup$ – Learnmore Nov 20 '16 at 3:06
  • $\begingroup$ @learnmore I suggest you have a closer look. The bound is not only correct, it is rather crude. Find an $n$ for which it is invalid and revert. $\endgroup$ – Mark Viola Nov 20 '16 at 4:10
  • $\begingroup$ @learnmore $2n-(n-1) =n+1$, whereupon dividing by $2n$ yields $1/2+1/(2n)$. So what on earth is the issue? $\endgroup$ – Mark Viola Nov 20 '16 at 4:18
  • $\begingroup$ Sorry,I am still having an issue;In the first line should not it be $2n(2n-1)(2n-2)\cdots (2n-(n-1)).n(n-1)\cdots 3.2.1$ $\endgroup$ – Learnmore Nov 20 '16 at 5:33
  • $\begingroup$ @learnmore Ah, yes, a typo. Great catch! I've edited accordingly. Much appreciative. -Mark $\endgroup$ – Mark Viola Nov 20 '16 at 5:39
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$$n!\geq\left (\frac{n}{4}\right )^{3n/4}\\\lim_{n\to\infty}\frac{\left (\frac{n}{4}\right )^{3n/4}}{n^{n/2}}=\lim_{n\to\infty}\frac{n^{n/4}}{4^{3n/4}}=\lim_{n\to \infty}\left(\frac{n}{64}\right)^{n/4}=\infty$$ The inequality comes since $4n!=1\cdots \underbrace{n\cdots 4n}_{3n}\geq \underbrace{n\cdots n}_{3n}=n^{3n}$

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  • $\begingroup$ That's a nice approach+1 $\endgroup$ – Learnmore Nov 20 '16 at 3:00
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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, $$ \lim_{n\to\infty}\frac{n!}{n^{n/2}}=\infty\tag{3} $$


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 $\ge1$.

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  • $\begingroup$ Is this the only way to solve this?Can you please provide some alternative way may be using inequalities etc. $\endgroup$ – Learnmore Nov 19 '16 at 18:05
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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 $\infty.$

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

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    $\begingroup$ The fact that $n!\ge \sqrt{n^n}$ does not imply alone that the limit of the ratio is infinite. $\endgroup$ – Mark Viola Nov 19 '16 at 17:54
  • $\begingroup$ What is the limit of (n! + 1) / n! ? The numerator is always larger than the denominator, so therefore the limit goes to infinity, right? That's your theory here. But that's absurd; in my example the sequence never even gets larger than 3. So where is the flaw in the argument? $\endgroup$ – Eric Lippert Nov 19 '16 at 19:40
  • $\begingroup$ @Dr.MV, Thanks for the correction. Please see my edit. $\endgroup$ – k170 Dec 27 '16 at 6:13
  • $\begingroup$ @EricLippert, please see my edit and thanks for the correction. $\endgroup$ – k170 Dec 27 '16 at 14:33

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