rational function limit involving factorials I posted something similar but someone edited the question and added the wrong equation, which gave irrelevant responses. 
Lim (2n-1)!/(2n)^n as n approach infinity. 
Any method, I would just like a starting, no full answer needed (but will be appreciated)!
 A: Hint. You may recall the Stirling formula
$$
n!\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n
$$ as $n$ is great, giving
$$
\frac{(2n-1)!}{(2n)^n}\sim\sqrt{2\pi (2n-1)}\left(\dfrac{2n-1}{e}\right)^{2n-1}\frac{1}{(2n)^n} \sim\frac{\sqrt{2\pi}}{e}\left(\dfrac{2n}{e^2}\right)^{n-1/2} 
$$
then your limit is $+\infty$ since $\displaystyle \dfrac{2n}{e^2} \longrightarrow +\infty$.
A: To eliminate the
messy $2n-1$,
write it as
$f(n)
=\frac{(2n)!}{(2n)^{n+1}}
$.
By Stirling,
$(2n)!
\approx \sqrt{2\pi (2n)} \frac{(2n)^{2n}}{e^{2n}}
= 2\sqrt{\pi n} \frac{(2n)^{2n}}{e^{2n}}
$.
Therefore
$f(n)
\approx \frac{2\sqrt{\pi n} \frac{(2n)^{2n}}{e^{2n}}}{(2n)^{n+1}}
= \frac{2\sqrt{\pi n} (2n)^{2n}}{(2n)^{n+1}e^{2n}}
= \frac{2\sqrt{\pi n} (2n)^{n-1}}{e^{2n}}
= \frac{2\sqrt{\pi n} (2n)^{n-1}}{e^2e^{2n-2}}
= \frac{2\sqrt{\pi n}}{e^2}\frac{ (2n)^{n-1}}{e^{2n-2}}
= \frac{2\sqrt{\pi n}}{e^2}\left(\frac{ 2n}{e^2}\right)^{n-1}
$
and the right side
pretty obviously gets
unbounded.
A: Hint: approximate $(2n-1)!$ using Stirling...
A: My suggestion is use the Stirling´s fórmula $n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$.
$\lim_{n \to \infty}\frac{(2n-1)!}{(2n)^n}\sim \lim_{n \to \infty}\frac{1}{(2n)^n}\sqrt{2\pi (2n-1)}(\frac{2n-1}{e})^{2n-1} $
$=\lim_{n \to \infty}\frac{1}{e^{2n-1}}\sqrt{2\pi(2n-1)}\frac{1}{(2n)^n}(2n-1)^n(2n-1)^{2n-1}$
$=\lim_{n \to \infty}\sqrt{2\pi(2n-1)}(\frac{2n-1}{2n})^n(\frac{2n-1}{e})^{2n-1}$
Now, note that,
$\lim_{n \to \infty}\sqrt{2\pi(2n-1)}=+\infty, \lim_{n \to \infty}(\frac{2n-1}{2n})^n=e^{-1/2} \ e  \ \lim_{n \to \infty}(\frac{2n-1}{e})^{2n-1}=+\infty$
hence,
$\lim_{n \to \infty}\frac{(2n-1)!}{(2n)^n}\sim\lim_{n \to \infty}\sqrt{2\pi(2n-1)}(\frac{2n-1}{2n})^n(\frac{2n-1}{e})^{2n-1}=+\infty$
A: Use  the formula $|a_{n+1} / a_n|$ 
