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

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

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

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