How to show $\frac12\cdot\frac34\cdot\frac56\cdots\frac{99}{100}<\frac{1}{12}$?

Question

How can I show that $$\frac12\cdot\frac34\cdot\frac56\cdots\frac{99}{100}<\frac{1}{12}?$$

Answer 1

For $m,n\in\mathbb{N}$ with $m\leq n$, let $S_{m,n}:=\prod_{i=m}^n\,\frac{2i-1}{2i}$. Then, $$S_{m,n}^2=\prod_{i=m}^n\,\left(\frac{2i-1}{2i}\right)^2 \leq \prod_{i=m}^n\,\left(\frac{2i-1}{2i}\right)\left(\frac{2i}{2i+1}\right)=\prod_{i=2m-1}^{2n}\,\frac{i}{i+1}=\frac{2m-1}{2n+1}\,.$$ Therefore, $S_{m,n}\leq\sqrt{\frac{2m-1}{2n+1}}$. In particular, $S_{4,50}\leq\sqrt{\frac{7}{101}}$. Now, $S_{1,50}=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot S_{4,50}$. Hence, $$S_{1,50}\leq\frac{15}{48}\sqrt{\frac{7}{101}}<\frac{1}{12}\,.$$

You can also show that $S_{m,n}\geq\sqrt{\frac{m-1}{n}}$, so $S_{1,50}\geq\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot\sqrt{\frac{4}{50}}=\frac{7}{64\sqrt{2}}$. This will give you $\frac{1}{13}<S_{1,50}<\frac{1}{12}$. Asymptotically, $S_{1,n}\sim\frac{1}{\sqrt{\pi n}}$. Indeed, $$\frac{1}{\sqrt{\pi\left(n+\frac12\right)}}<S_{1,n}<\frac{1}{\sqrt{\pi n}}\,.$$

Answer 2

Assuming that you know factorials and their approximations, rewrite $$A=\frac12\times\frac34\times\frac56\times\cdots\times\frac{99}{100}=\frac12\times\frac22\times\frac34\times\frac44\times\frac56\times\frac66\times\cdots\times \frac{99}{100}\times \frac{100}{100}$$ $$A=\frac{1}{4^{50}}\frac{1\times2\cdots\times 100}{(1\times2\cdots\times 50)^2}=\frac{1}{4^{50}}\frac{100!}{(50!)^2}$$ At this point, we can use Stirling approximation of the factorial $$n!\approx \sqrt{2\pi n} \big(\frac n e\big)^n$$ So $$A_n=\frac{(2n)!}{4^n (n!)^2}\approx \frac{1}{\sqrt{\pi n}}$$ For the problem, use $n=50$.