Limit of sequence. with Factorial

Question

Can't find the limit of this sequence : $$\frac{3^n(2n)!}{n!(2n)^n}$$ tried to solve this using the ratio test buy failed... need little help

Answer 1

What's the problem with the ratio test?:

$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!\color{red}{3^{n+1}}}{(n+1)!(\color{green}{2}(n+1))^{n+1}}\frac{n!(\color{green}{2}n)^n}{(2n)!\color{red}{3^n}}=\frac{(2n+1)\cdot3}{n+1}\left(\frac{n}{n+1}\right)^n\xrightarrow[n\to\infty]{}\frac6e>1$$

and thus...

Answer 2

Hint: use Stirling formula, $n! \simeq \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$

Answer 3

As tenpercent said, since $n! \sim \sqrt{2\pi n} (n/e)^n$,

$\begin{align} \frac{3^n(2n)!}{n!(2n)^n} &\sim \frac{3^n\sqrt{2\pi 2n} (2n/e)^{2n}}{\sqrt{2\pi n} (n/e)^n(2n)^n}\\ &= \frac{3^n2^{2n}n^{2n}\sqrt{4\pi n} /e^{2n}}{\sqrt{2\pi n}2^n n^{2n}/e^n}\\ &= \frac{3^n2^{n}\sqrt{2} }{e^n}\\ &=\left( \frac{6}{e}\right)^n \sqrt{2} \end{align} $