Evaluate $\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}$

Question

$$\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}$$

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

Answer 1

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 $$ \dfrac{a_{n+1}}{a_n} = \dfrac{(2(n+1))!}{(n+1)^{n+1}{(n+1)!}} \ \dfrac{n^n{n!}}{(2n)!} = \dfrac{(2n+2)(2n+1)}{(n+1)(n+1)} \ \left(\dfrac{n}{n+1}\right)^n \to \dfrac{4}{e} $$

Answer 2

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} $

Answer 3

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 , $\displaystyle a_n=\frac{(2n)!}{n^2.n!}$.