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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.

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juantheron
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    Note that $\binom{2n}{n}$ must be larger than the average of all the $\binom{2n}{k}$, which is $\frac{4^n}{2n+1}$. This isn't quite as good as your bound, but it's definitely good enough, and I think it's a lot more straightforward than your Cauchy-Schwarz argument... – Micah Oct 06 '16 at 16:07
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    Evaluate a $\underline{limit}$ $\texttt{without}$ using $\underline{limit}$ ?. – Felix Marin Oct 06 '16 at 21:13
  • There are literally a handful of duplicates of this question all over the site. Asking not to use "limit as a sum" doesn't make sense given that you are using Cauchy-Schwarz. Be more precise about what you are not allowing. – heropup Oct 06 '16 at 21:41

3 Answers3

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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 $4$.

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Cm7F7Bb
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The upper bound is way more trivial than the lower bound through the CS inequality: $$ \binom{2n}{n}\leq \sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n} = 4^n.$$

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

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}

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