Inequality $\binom{2n}{n}\leq 4^n$

Question

I would like to prove the following inequality, for $n=0,1,2,...$, $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.

Answer 1

$\displaystyle 4^n=(1+1)^{2n} =\sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}$

Answer 2

A bit more can be proven with a bit more work. For $k\ge0$, we have the inequality $$ \begin{align} \left(\frac{k+\frac12}{k+1}\right)^2 &=\frac{k^2+k+\frac14}{k^2+2k+1}\\ &\le\frac{k+1}{k+2}\tag{1} \end{align} $$ because cross-multiplication gives $k^3+3k^2+\frac94k+\frac12\le k^3+3k^2+3k+1$.

Using $(1)$ yields $$ \begin{align} \frac{\binom{2k+2}{k+1}}{\binom{2k}{k}} &=4\frac{k+\frac12}{k+1}\\ &\le4\sqrt{\frac{k+1}{k+2}}\tag{2} \end{align} $$ Multiplying $(2)$ for $k=0$ to $k=n-1$, we get $$ \boxed{\bbox[5px]{\displaystyle\binom{2n}{n}\le\frac{4^n}{\sqrt{n+1}}}}\tag{3} $$

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As Olivier Oloa comments, Stirling's Formula tells us that $$ \lim_{n\to\infty}\binom{2n}{n}\frac{\sqrt{\pi n}}{4^n}=1\tag{4} $$ In fact, using inequalities similar to $(2)$, in this answer, it is shown that $$ \boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}}}\tag{5}$$

Answer 3

Let $n \in \mathbb{N}.$ You may write

$$ 4^n=2^{2n} =(1+1)^{2n}= \sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}. $$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{2n \choose n}\leq 4^{n}:\ {\large ?}}$

\begin{align} \binom{2n}{n}&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic} \end{align}

\begin{align} \color{#66f}{\large\verts{\binom{2n}{n}}}&= \verts{\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}} \,{\dd z \over 2\pi\ic}} \\[5mm] \leq & \oint_{\verts{z}\ =\ 1}\verts{\pars{1 + z}^{2n}} \,{\verts{\dd z} \over 2\pi}<\pars{1 + 1}^{2n}=\color{#66f}{\LARGE 4^{n}} \end{align}