Prove that
$$\lim_{n\to\infty}{\root{n}\of{\frac{(2n)!}{(n!)^2}}}=+\infty$$
(Edit: as noted in the answers the actual limit is 4; I stand corrected.)
The argument of the $n$th root goes to infinity since it is larger than $n$, also it grows faster than than $n^k$ for any $k$... but taking the $n$th root of any of these would make the limit 1.
Is there a simple argument or should I use some trick...?
Answer
Set $a_n=\binom{2n}{n}$. Then if the quotient sequence $\frac{a_{n+1}}{a_n}$ converges to a limit $L$, the root sequence $\sqrt[n]{a_n}$ has the same limit.
As
$$
\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}=4·\frac{1+\frac1{2n}}{1+\frac1n}
$$
does indeed converge, both sequences have the same limit $L=4$.
Bonus info: See also Newton's binomial series where
$$
(1-x)^{-\frac12}
=\sum_{k=0}^\infty\binom{-\frac12}k· (-x)^k
=\sum_{k=0}^\infty\binom{2k}{k}·\left(\frac x4\right)^k
$$
which has, as all binomial series, $R=1$ as radius of convergence, giving also the limit of the root sequence as $4$ by the theorem of Cauchy-Hadamard on the sharpness of the root formula for that radius.
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