# Elementary central binomial coefficient estimates

1. How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?

2. Does anyone know any better elementary estimates?

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For (2), Stirling's approximation yields ${2n\choose n}\sim (\pi n)^{-1/2} 4^n$, and you might look at some of the other formulas for sharper bounds. –  anon Aug 19 '11 at 20:49
@anon The Central Limit Theorem shows that eventually the Stirling approximation must be an upper bound (and it actually is for all positive $n$). A lower bound is obtained the same way by multiplying the upper bound by $\exp(-1/(4n))$. Because this is $1 + O(1/n)$ for large $n$, it's better than the OP's bounds whose ratio equals $O(\sqrt{4/3})$. –  whuber Aug 19 '11 at 20:55
@whuber: 1,000,000=O(sqrt(4/3)). en.wikipedia.org/wiki/Big_O_notation#Formal_definition –  Did Aug 20 '11 at 12:32

Here are some crude bounds:

$${1\over 2\sqrt{n}}\leq {2n\choose n}{1\over 2^{2n}}\leq{3\over4\sqrt{n+1}},\quad n\geq1.$$

We begin with the product representations $${2n\choose n}{1\over 2^{2n}}={1\over 2n}\prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)=\prod_{j=1}^n\left(1-{1\over2j}\right),\quad n\geq1.$$

From $$\prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)^{\!\!2}=\prod_{j=1}^{n-1}\left(1+{1\over j}+{1\over 4j^2}\right)\geq \prod_{j=1}^{n-1}\left(1+{1\over j}\right)=n,$$ we see that $$\left({2n\choose n}{1\over 2^{2n}} \right)^{2} = {1\over (2n)^2}\, \prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)^{\!\!2} \geq {1\over 4n^2}\, n ={1\over 4n},\quad n\geq1.$$ so by taking square roots, ${2n\choose n}{1\over 2^{2n}}\geq \displaystyle{1\over 2\sqrt{n}}.$

On the other hand, $$\prod_{j=1}^{n}\left(1+{1\over 2j}\right) \left(1-{1\over 2j}\right) = \prod_{j=1}^{n}\left(1-{1\over 4j^2}\right)\leq {3\over 4},$$ so that (using the lower bound above), we have $${2n\choose n}{1\over 2^{2n}}=\prod_{j=1}^n\left(1-{1\over2j}\right)\leq{3\over4\sqrt{n+1}}.$$

Alternatively, multiplying the different representations we get $$n\left[{2n\choose n}{1\over 2^{2n}}\right]^2={1\over 2}\prod_{j=1}^{n-1}\left(1-{1\over4j^2}\right) \,\left(1-{1\over 2n}\right).$$ It's not hard to show that the right hand side increases from $1/4$ to $1/\pi$ for $n\geq 1$.

Edit: You can get better bounds if you know Wallis's formula:

$$2n\left[{2n\choose n}{1\over 4^n}\right]^2={1\over 2}{3\over 2}{3\over 4}{5\over 4}\cdots {2n-1\over 2n-2}{2n-1\over 2n}={1\over 2}\prod_{j=2}^n\left(1+{1\over 4j(j-1)}\right)$$

$$(2n+1)\left[{2n\choose n}{1\over 4^n}\right]^2={1\over 2}{3\over 2}{3\over 4}{5\over 4}\cdots {2n-1\over 2n-2}{2n-1\over 2n}{2n+1\over 2n}=\prod_{j=1}^n\left(1-{1\over 4j^2}\right)$$

By Wallis's formula, both middle expressions converge to ${2\over \pi}$. The right hand side of the first equation is increasing, while the right hand side of the second equation is decreasing. We conclude that

$${1\over\sqrt{\pi(n+1/2)}}\leq {2n\choose n}{1\over 4^n}\leq {1\over\sqrt{\pi n}}.$$

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A way to get explicit bounds via Stirling's approximation is to use the following more precise formulation: $$n! = \sqrt{2\pi n} \left( \frac{n}{e} \right)^n e^{\alpha_n}$$ where $\frac{1}{12n+1} < \alpha_n < \frac{1}{12n}$.

With this one arrives at $$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} e^{\lambda_n}$$ where $\frac{1}{24n+1} - \frac{1}{6n} < \lambda_n < \frac{1}{24n} - \frac{2}{12n+1}$.

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@ThomasAndrews Ahh yes thank you for pointing that out. –  Ragib Zaman Mar 14 '13 at 0:21

You can get an even more precise answer than those already provided by using more terms in the Stirling series. Doing so yields, to a relative error of $O(n^{-5})$,

$$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1 - \frac{1}{8n} + \frac{1}{128n^2} + \frac{5}{1024n^3} - \frac{21}{32768 n^4} + O(n^{-5})\right).$$

To the same relative error of $O(n^{-5})$, the Stirling series is $$n!=\sqrt{2\pi n}\left({n\over e}\right)^n \left( 1 +{1\over12n} +{1\over288n^2} -{139\over51840n^3} -{571\over2488320n^4} + O(n^{-5}) \right).$$

Then we have $$\binom{2n}{n} = \frac{(2n)!}{n! n!} = \frac{4^n}{\sqrt{\pi n}} \frac{1 + \frac{1}{12(2n)} + \frac{1}{288(2n)^2} - \frac{139}{51840(2n)^3} - \frac{571}{2488320(2n)^4} + O(n^{-5})}{\left(1 + \frac{1}{12n} + \frac{1}{288n^2} - \frac{139}{51840n^3} - \frac{571}{2488320n^4} + O(n^{-5})\right)^2}.$$

Crunching through the long division with the polynomial in $\frac{1}{n}$ (which Mathematica can do immediately with the command Series[expression, {n, ∞, 4}]) yields the expression for $\binom{2n}{n}$ at the top of the post.

$\displaystyle\frac{4^n}{\sqrt{\pi n}}\left(1-\frac1{8n}+O\left(\frac1{n^2}\right)\right)$ matches the upper bound gotten in my answer. I remember computing this asymptotic series at one point. (+1) –  robjohn Sep 16 '14 at 10:38
For $n\ge0$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\le\frac{n+\frac13}{n+\frac43}\tag{1} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2} \end{align} Inequality $(2)$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}$$ For $n\ge0$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\ge\frac{n+\frac14}{n+\frac54}\tag{4} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5} \end{align} Inequality $(5)$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}$$ Note that the formula in $(3)$, which is decreasing, is bigger than the formula in $(6)$, which is increasing. Their ratio tends to $1$; therefore, they tend to a common limit, $L$.
Note that \begin{align} \frac{\binom{2n}{n}}{4^n}\sqrt{n} &=\frac{2n-1}{2n}\frac{2n-3}{2n-2}\cdots\frac12\sqrt{n}\\ &=\frac{\Gamma(n+\frac12)/\Gamma(\frac12)}{\Gamma(n+1)/\Gamma(1)}\sqrt{n}\tag{7} \end{align} Using Gautschi's Inequality, we have $$\sqrt{\frac{n}{n+1}}\le\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}\sqrt{n}\le1\tag{8}$$ Therefore, $(7)$ and $(8)$ yield \begin{align} L &=\lim_{n\to\infty}\frac{\binom{2n}{n}}{4^n}\sqrt{n}\\ &=\frac{\Gamma(1)}{\Gamma(\frac12)}\\ &=\frac1{\sqrt\pi}\tag{9} \end{align} Combining $(3)$, $(6)$, and $(9)$, we get $$\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}}}\tag{10}$$ If we know that $n\ge n_0$, we can replace $\frac13$ in $(1)-(3)$ and $(10)$ by $\frac14+\frac1{16n_0+12}$. This gives us the asymptotic expansion $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}=\frac{4^n}{\sqrt{\pi n}}\left(1-\frac1{8n}+O\left(\frac1{n^2}\right)\right)}}\tag{11}$$
@ByronSchmuland: Thanks! I came up with the idea of the estimates $(3)$ and $(6)$ while working on this answer, then expanded on it when I found this question. –  robjohn Sep 16 '14 at 15:05