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
$.