Evaluate $\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}$ $$\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}$$
It is a sequence and n is natural
It looks like I should use $\lim_{n\to \infty}{\sqrt[n]{a_n}}=x$ but I don't know how. Does it mean that $a_n=\frac{(2n)!}{n^n\times{n!}}$ and then do it from there or is $a_n=\sqrt[n]\frac{(2n)!}{n^n\times{n!}}$
I have never used this before and I am not sure what to do
 A: Let $a_n = \dfrac{(2n)!}{n^n{n!}}$. Then
$\displaystyle
\lim_{n\to \infty}{\sqrt[n]{a_n}}
=
\lim_{n\to \infty}\dfrac{a_{n+1}}{a_n}
$
if this limit exists.
Now
$$
\dfrac{a_{n+1}}{a_n}
=
\dfrac{(2(n+1))!}{(n+1)^{n+1}{(n+1)!}}
\
\dfrac{n^n{n!}}{(2n)!}
=
\dfrac{(2n+2)(2n+1)}{(n+1)(n+1)}
\
\left(\dfrac{n}{n+1}\right)^n
\to
\dfrac{4}{e}
$$
A: Since
$\lim_{n \to \infty} (n!)^{1/n}
\approx (n/e)
$,
$\begin{array}\\
\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}
&=\lim_{n\to \infty}{\frac{((2n)!)^{1/n}}{n(n!)^{1/n}}}\\
&\approx\frac{\lim_{n\to \infty}((2n)!)^{1/n)}}{n(n/e)}\\
&=\frac{\lim_{n\to \infty}(((2n)!)^{1/(2n)})^2}{n(n/e)}\\
&\approx\frac{(2n/e)^2}{n(n/e)}\\
&=\frac{4}{e}
\end{array}
$
You can make it more general
and use
$\lim_{n \to \infty} ((kn)!)^{1/n}
=\lim_{n \to \infty} (((kn)!)^{1/(kn)})^k
=(kn/e)^k
$
to show that
$\lim_{n\to \infty}\sqrt[n]{\frac{(an)!}{n^{cn}(bn)!}}$
exists only when
$a \le b+c$
and is zero when
$a < b+c$
and is
$\frac{a^a}{b^b}e^{b-a}
$
when
$a = b+c$.
The OP is
$a=2, b = 1, c=1$.
$\begin{array}\\
\lim_{n\to \infty}\sqrt[n]{\frac{(an)!}{n^{cn}(bn)!}}
&\approx\lim_{n\to \infty}\frac{(an/e)^a}{n^c(bn/e)^b}\\
&=\lim_{n\to \infty}\frac{a^a}{b^b}e^{b-a}n^{a-b-c}\\
\end{array}
$
and this exists as stated.
Note:
If you want to
make this more rigorous,
and use
$\frac{(n!)^{1/n}}{n}
\approx \frac1{e}
$,
and similarly use
$\frac{(kn)!^{1/n}}{n^k}
=\frac{((kn)!^{1/(kn)})^k}{n^k}
\approx\frac{(kn/e)^k}{n^k}
=(k/e)^k
$
you can write
$\begin{array}\\
\sqrt[n]{\frac{(an)!}{n^{cn}(bn)!}}
&=\frac{\frac{((an)!)^{1/n}}{n^a}}{\frac{((bn)!)^{1/n}}{n^b}}
\frac{n^a}{n^{c}n^b}\\
&\approx\frac{(a/e)^a}{(b/e)^b}n^{a-b-c}\\
&=\frac{a^a}{b^b}e^{b-a}n^{a-b-c}\\
\end{array}
$
A: Hint : Use Cauchy's second limit theorem.
If $\displaystyle \lim_n\frac{a_{n+1}}{a_n}=l$ , then $\displaystyle \lim_na_n^{1/n}=l$. Where , $\displaystyle a_n=\frac{(2n)!}{n^2.n!}$.
