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

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.

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

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 $$\boxed{\bbox[5px]{\displaystyle\binom{2n}{n}\le\frac{4^n}{\sqrt{n+1}}}}\tag{3}$$

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 $$\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}}}\tag{5}$$

• Nice inequality, especially knowing that the true order is $$\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi}\sqrt{n}}.$$(+1) – Olivier Oloa Sep 14 '14 at 21:37
• @OlivierOloa: We actually have that $$\binom{2n}{n}\frac{\sqrt{\pi n}}{4^n}\nearrow1$$ which means that $\frac{4^n}{\sqrt{\pi n}}$ is an overestimate as well as an asymptotic approximation. Come to think of it, I can prove this using a similar argument and Stirling's Formula. Time for an edit :-) – robjohn Sep 14 '14 at 23:24

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

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

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

\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}}\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}