Evaluation of $\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}$ 
Evaluation of $$\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}$$
without using Limit as a sum and stirling Approximation.

$\bf{My\; Try:}$ Using $$\binom{2n}{n} = \sum^{n}_{r=0}\binom{n}{r}^2$$
Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\left[\sum^{n}_{r=0}\binom{n}{r}^2\right]\cdot \left[\sum^{n}_{r=0}1\right]\geq \left(\sum^{n}_{r=0}\binom{n}{r}\right)^2 = 2^{2n} = 4^n$$
So $$\frac{4^n}{n+1}<\sum^{n}_{r=0}\binom{n}{r}^2=\binom{2n}{n}$$
But i did not understand how can i calculate upper bound such that i can apply the Squeeze theorem.
 A: We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then
$$
\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}
$$
(see this answer).
We have that
$$
\frac{{2n+2\choose n+1}}{{2n\choose n}}=\frac{(2n+2)!}{((n+1)!)^2}\frac{(n!)^2}{(2n)!}=\frac{(2n+2)(2n+1)}{(n+1)^2}\to4
$$
as $n\to\infty$. Hence, the limit is $4$.
A: The upper bound is way more trivial than the lower bound through the CS inequality:
$$ \binom{2n}{n}\leq \sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n} = 4^n.$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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By "$\textsf{using limits}$", the Stoltz-Ces$\grave{a}$ro Theorem yields:

\begin{align}
\lim_{n \to \infty}{1 \over n}\,\ln\pars{{2n \choose n}} & =
\lim_{n \to \infty}{\ln\pars{{2\bracks{n + 1} \choose n + 1}} -
\ln\pars{{2n \choose n }} \over \pars{n + 1} - n} =
\lim_{n \to \infty}\ln\pars{{2n + 2 \choose n+1} \over {2n \choose n}}
\\[5mm] & =
\lim_{n \to \infty}\ln\pars{\bracks{2n + 2}\bracks{2n + 1} \over
\bracks{n + 1}\bracks{n + 1}} = \ln\pars{4}
\end{align}

$$
 \color{#f00}{\lim_{n \to \infty}{2n \choose n}^{1/n}} = \color{#f00}{4}
$$
