Limit of sequences: $\lim \frac{(2n)!}{(n!)^2} $ Verify if the sequence
$$\frac{(2n)!}{(n!)^2}$$
converges.
My attempt:
$$\frac{(2n)!}{(n!)^2} = \frac{(2n)(2n-1)...(n+1)}{n.(n-1)...1} \geq \frac{(n+1)^n}{n!}
$$
Maybe it is easier to show that this last sequence diverges.
Thanks!
 A: You are very close. Can you see that
$$\frac{(2n)!}{(n!)^2} = \frac{(2n)(2n-1)...(n+1)}{n.(n-1)...1} \geq 2^n$$
A: Hint: Note that
$$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$$
Then the sequence diverges.
A: $$\frac{(2n)!}{(n!)^2}=\binom{2n}{n}=4\left(1-\frac{1}{2n}\right)\cdot\binom{2n-2}{n-1}$$
hence, since $1-x\geq \exp\frac{x}{x-1}$:
$$\begin{eqnarray*}\binom{2n}{n}&=&4^n \prod_{j=1}^{n}\left(1-\frac{1}{2j}\right)\geq 4^n\exp\left(-\sum_{j=1}^n \frac{1}{2j-1}\right)\\   &\geq&4^n\exp\left(-1-\frac{1}{2}\log n\right)=\frac{4^n}{e\sqrt{n}}.\end{eqnarray*}$$
A: $$\frac{(2n)!}{(n!)^2} = \frac{(2n)(2n-1)...(n+1)}{n.(n-1)...1} \geq \frac{(n+1)^n}{n!} \geq \frac{n^n}{n!}
$$
If you prove that the sequnece $a_n = \frac{n!}{n^n}$ converges to 0, then you have proven that $\lim_{n \to \infty}\frac{n^n}{n!} = \infty$ and thus your original sequence also goes to $\infty$. 
You can use the monotone convergence theorem. First prove that the sequence is decreasing, 
$$
\frac{n!}{n^n} \geq \frac{(n+1)!}{(n+1)^{n+1}} \Leftrightarrow \left(\frac{n+1}{n}\right)^n \geq 1 \Leftrightarrow \left(1 + \frac{1}{n}\right)^n \geq 1
$$
Then because the sequence is bounded, $\frac{n!}{n^n}\geq 0$, it must converge. So a limit exists, lets call it $\lim_{n \to \infty}\frac{n!}{n^n} =a$, then $\lim_{n \to \infty} a_n = \lim_{n \to \infty}a_{n+1} = a $. We can express $a_{n+1}$ with $a_n$
$$
a_{n+1} = a_n\left(\frac{n}{n+1} \right)^n
$$
It the limit as $n \to \infty$ this becomes
$$
a = a\frac{1}{e}
$$
So $a$ must be $0$. 
A: Just use the result

If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = a  $ and $|a|>1$ then $\lim_{n\to \infty} = \infty$. 

In your case 

$$ \frac{a_{n+1}}{a_n} = \frac{2(n+1)}{n+1}. $$

I think you can finish it.
A: Using the quotient test,
$$
\lim_{n\to\infty}\frac{\frac{((n+1)!)^2}{(2(n+1))!}}{\frac{(n!)^2}{(2n)!}}=
\lim_{n\to\infty}\frac{((n+1)!)^2}{(2n+2)!}\frac{(2n)!}{(n!)^2}=
\lim_{n\to\infty}\frac{(n+1)^2}{(2n+1)(2n+2)}=\frac14
$$
the series $\sum\frac{(n!)^2}{(2n)!}$ is convergent. This implies
$\lim_{n\to\infty}\frac{(n!)^2}{(2n)!}=0$.
