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
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See https://en.wikipedia.org/wiki/Central_binomial_coefficient. – lhf Jul 11 '16 at 12:16
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1Some proofs there – Zubzub Jul 11 '16 at 12:17
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Also http://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates – Jul 11 '16 at 13:14
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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.
lhf
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1@Ihf in second line is it "$2n$" or "$n$"?..can you elaborate the last line when n tends to infinity? how you derived that term? – ViX28 Jul 11 '16 at 12:33
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@ViX28: It goes to infinity as $n$ goes to infinity, but grows more slowly than $2^n$, i.e. is $o(2^n)$ – Henry Jul 11 '16 at 13:07
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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:
$$\forall n\geq 1,\qquad \frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}+\frac{1}{32n}\right)}}.\tag{2}$$
Jack D'Aurizio
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