How to solve the following limit using mathematics Stirling $\lim\limits_{n\to \infty}\frac{n!}{n^ne^{-n}\sqrt{2\pi n}}=1$

Question

How to solve the following limit using mathematics Stirling $\lim\limits_{n\to \infty}\frac{n!}{n^ne^{-n}\sqrt{2\pi n}}$.

a) $\lim\limits_{n\to \infty}\frac{n!e^n}{n^{n+1/2}}$.

b) $\lim\limits_{n\to \infty}\frac{(2n)!}{e^{-2n}(2n)^{2n}\sqrt{n}}$.

c) $\lim\limits_{n\to \infty}\frac{(2n)!\sqrt{n}}{n!^24^{n}}$.

d) $\lim\limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}$.

For a), $\frac{n!e^n}{n^{n+1/2}}$=$\frac{\sqrt{2\pi}}{\sqrt{2\pi}}\frac{n!}{n^{n} e^{-n} \sqrt{n}}$. Therefore the limit is $\sqrt{2\pi}$

For b) and c), I don't know how we can modify $(2n)!$

For d), $\frac{\sqrt[n]{n!}}{n}=\frac{\sqrt[n]{n}\sqrt[n]{(n-1)!}}{n}$. But then I don't know how to proceed?

Could someone help?

Answer 1

First of all, you need to state Stirling's theorem in its proper form: $\lim\limits_{n\to \infty}\frac{n!}{n^ne^{-n}\sqrt{2\pi n}} =1 $.

By multiplying by $\sqrt{2\pi}$, you can restate this as $\lim\limits_{n\to \infty}\frac{n!}{n^ne^{-n}\sqrt{ n}} =\sqrt{2\pi} $.

To answer (b), replace $n$ by $2n$ above to get $\sqrt{2\pi} =\lim\limits_{n\to \infty}\frac{(2n)!}{(2n)^{2n}e^{-2n}\sqrt{ 2n}} $, or, multiplying by $\sqrt{2}$, $2\sqrt{\pi} =\lim\limits_{n\to \infty}\frac{(2n)!}{(2n)^{2n}e^{-2n}\sqrt{ n}} $.

Combining the results for $n!$ and $(2n)!$, which is a standard technique to estimate $\binom{2n}{n}$, and using $f(n) \approx g(n)$ to mean $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 1 $, we have $n! \approx n^ne^{-n}\sqrt{2\pi n} $ and $(2n)! \approx (2n)^{2n}e^{-2n}2\sqrt{\pi n} $

so that

$\begin{array}\\ \frac{(2n)!}{n!^2} &\approx \frac{(2n)^{2n}e^{-2n}2\sqrt{\pi n}}{(n^ne^{-n}\sqrt{2\pi n})^2}\\ &= \frac{2^{2n}n^{2n}e^{-2n}2\sqrt{\pi n}}{n^{2n}e^{-2n}2\pi n}\\ &= \frac{4^{n}}{\sqrt{\pi n}}\\ \text{or}\\ \frac{(2n)!\sqrt{\pi n}}{n!^24^{n}} &\approx 1\\ \end{array} $

so that $\lim\limits_{n\to \infty}\frac{(2n)!\sqrt{n}}{n!^24^{n}} =\frac1{\sqrt{\pi}} $.

For the last one, since $\lim\limits_{n\to \infty}\frac{n!}{n^ne^{-n}\sqrt{2\pi n}} =1 $, taking the $n$-th root, this becomes

$\begin{array}\\ 1 &=\lim\limits_{n\to \infty}\left(\frac{n!}{n^ne^{-n}\sqrt{2\pi n}}\right)^{1/n}\\ &=\lim\limits_{n\to \infty}\left(\frac{n!} {n^ne^{-n}}\right)^{1/n}\frac1{\sqrt{2\pi n}^{1/n}}\\ &=\lim\limits_{n\to \infty}\frac{n!^{1/n}e} {n} \qquad\text{since } n^{1/n}\to 1 \text{ and } c^{1/n}\to 1\\ \end{array} $

so

$\lim\limits_{n\to \infty}\frac{n!^{1/n}} {n} = e $.