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  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
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@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
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3 Answers 3

up vote 14 down vote accepted

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
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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.

See also Problem 9.60 in Concrete Mathematics (2nd edition).

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