Estimate for the binomial coefficients and bounds from below for the Beta function

Question

Let $n\ge p\in \mathbb N$ and let $\binom{n}{p}$ be the binomial coefficient. I believe that $$ \binom{n}{p}\le 2^n\sqrt\frac{2}{π n}. $$ Question: is that true? Of course I would like it as a non-asymptotic result, valid for all integers $n,p$. A related estimate would be a bound from below for the Beta function with $$ B(x,y)\ge \frac{\sqrt{(y-1)(x-1)}}{2^{x+y-1}}\sqrt\frac{π(x+y-1)}{2},\quad x,y\ge 1. $$

Answer 1

There is an elegant way to get the bounds for the binomial coefficients, using that the middle binomial coefficient is the largest.

First consider the even case, and put $$ a_n = \binom{2n}{n} \frac{\sqrt{2n}}{4^n}. $$ By Stirling's formula it is clear that $a_n \to \sqrt{2/\pi}$ as $n\to \infty$ and we want the inequality $a_n < \sqrt{2/\pi}$ for all $n$. This holds because the sequence $a_n$ is monotone increasing, which we may see by computing $$ \frac{a_{n+1}}{a_n} = \frac{\sqrt{n+1}}{\sqrt{n}} \frac{(2n+1)}{2(n+1)} = \Big( 1+ \frac{1}{4n^2+4n} \Big)^{\frac 12} > 1. $$

Now consider the odd case, and put $$ b_n = \binom{2n+1}{n} \frac{\sqrt{2n+1}}{2^{2n+1}}. $$ Once again it is enough to show that $b_n$ is monotone increasing, and again we compute readily that $$ \frac{b_n}{b_{n-1}} = \frac{\sqrt{2n+1}}{\sqrt{2n-1}} \frac{(2n+1)}{2(n+1)} = \Big( \frac{(2n+1)^3}{4(2n-1)(n+1)^2} \Big)^{\frac 12} > 1. $$

Answer 2

What do you get if you simply use a fairly precise version of Stirling's formula for the three factorials in $\binom{n}{p}$? For example, try using $$ \sqrt{2\pi}\cdot n^{n+1/2}\cdot e^{-n+1/(12n+1)} < n! < \sqrt{2\pi}\cdot n^{n+1/2}\cdot e^{-n+1/12n}. $$ And if that's not good enough, there are better estimates available. If I have a chance later, I'll try to work out the details, unless you do it first.

Answer 3

The estimate you state holds and is standard. Various Internet references on this exist, do a search for binomial coefficients.

For small $p$, one can approximate the coefficient with $(en/p)^p$ and $(e(n-p)/p)^p$, and with care one can prove for which $n$ and $p$ these strictly bound the coefficient. For $ \mid n-2p \mid \leq \sqrt{4\pi n}$, your posted bound is easiest and suffices for most applications. If you need really fine control, it may be best to estimate a product of a few terms of $(n-k)/k$ by an appropriate power of 2 and use that.

For $p$ not small and not close to $n/2$, the bound you choose may depend on the application. Most of the "mass" of the binomial distribution is centered around $n/2$, and likely good estimates for the tails are also found with an Internet search.

Gerhard "Found Lots On The Internet" Paseman, 2018.07.05.