Bound on the value of $\binom{n }{n/2}$

Question

I know the value of $\binom{n}{r}$ is maximum for $r=n/2$ if $n$ is even. I am in need to calculate the value of $\binom{n}{n/2}$. \begin{align*} \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\ldots+\binom{n}{n/2}+\ldots,+\binom{n}{n}=2^n \end{align*} What is the value of $\binom{n}{n/2}$? any idea how to calculate it. I think it is order of $2^ n$. Please help me regarding this

Answer 1

The central term is the largest one and so $2^n \le (n+1){{n} \choose {n/2}}$. Therefore $$ \frac{2^n}{n+1} \le {{n} \choose {n/2}} \le 2^n $$

It is true that $$ {n \choose {n/2}}\sim {\frac {2^{n}}{{\sqrt {\pi {n/2}}}}}{\text{ as }}n\rightarrow \infty $$ This follows from Wallis' product and from Stirling's approximation.

Answer 2

You my provide tight bounds for a central binomial coefficient through Stirling's inequality.
For instance,

$$-\frac{1}{64n\left(n+\frac{1}{2}\right)}\leq \log\,\left(\binom{2n}{n}\frac{\sqrt{\pi\left(n+\frac{1}{4}\right)}}{4^n}\right)\leq -\frac{1}{64n(n+1)} \tag{1}$$

holds for every $n\geq 1$. That gives that $\binom{2n}{n}$ essentially behaves like $\frac{4^n}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}.$
An even more accurate approximation is:

$$\forall n\geq 1,\qquad \frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}+\frac{1}{32n}\right)}}.\tag{2}$$