TITLE: Evaluation of $\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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