# inequality $\prod\limits_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$

$n$ is a positive integer, then

$$\prod_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$$

with mathematical induction, we can prove this.

But I would love to find a wonderful method without mathematical induction

Thank you!

• Can one refer to $\binom{2n}{n}$? – André Nicolas Jun 7 '14 at 7:28
• @AndréNicolas Yes, sure – 2016 Jun 7 '14 at 7:28
• math.stackexchange.com/questions/58560/… – user940 Jun 7 '14 at 15:37
• This question gives the better upper bound $\frac1{\sqrt{\pi n}}$. – user21467 Jun 7 '14 at 15:37

This is a beautiful inequality and, if you know or are allowed to use the $\Gamma$ function, the lhs simplifies to $$\prod_{k=1}^n\frac{2k-1}{2k}=\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (n+1)}$$ For large values of $n$, an asymptotic development is given by $$\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (n+1)}=\frac{\sqrt{\frac{1}{n}}}{\sqrt{\pi }}-\frac{\left(\frac{1}{n}\right)^{3/2}}{8 \sqrt{\pi }}+\frac{\left(\frac{1}{n}\right)^{5/2}}{128 \sqrt{\pi }}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$ and the difference between lhs and rhs is found to always be negative and to increase to $0^-$. For the difference, we have $$rhs-lhs=\left(\frac{1}{\sqrt{\pi }}-\frac{1}{\sqrt{3}}\right) \sqrt{\frac{1}{n}}-\frac{\left(\frac{1}{n}\right)^{3/2}}{8 \sqrt{\pi }}+\frac{\left(\frac{1}{n}\right)^{5/2}}{128 \sqrt{\pi }}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$

Building on André Nicolas' idea, you may rewrite $\prod_{k=1}^n\frac{2k-1}{2k}$ as

$$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdots\frac{2n-1}{2n}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots (2n-1)2n}{(2\cdot 4 \cdot 6\cdot 8\cdots 2n)^2}=\frac{(2n)!}{(2^n\cdot 1\cdot 2\cdot 3\cdots n)^2}=\frac{1}{4^n}\binom{2n}{n}.$$

Now, $\binom{2n}{n}$ is the central binomial coefficient. Its asymptotic is both well-known and easily determined as

$$\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}},$$

so that asymptotically,

$$\prod_{k=1}^n \frac{2k-1}{2k}\sim \frac{1}{\sqrt{\pi n}}<\frac{1}{\sqrt{3n}}\quad(\pi=3.1416...).$$

Moreover, wikipedia also lists a property of central binomial coefficients which directly proves your result, namely,

$$\binom{2n}{n}\le \frac{4^n}{\sqrt{3n+1}}<\frac{4^n}{\sqrt{3n}}.$$