Prove that $\frac{1}{15}<\frac{1}{2}*\frac{3}{4}* \dots *\frac{99}{100}<\frac{1}{10}$

Question

Prove that $\frac{1}{15}<\frac{1}{2}*\frac{3}{4}* \dots *\frac{99}{100}<\frac{1}{10}$

My attempt:If we name the value $A$ we have:

$A^2<\left(\frac{1}{2}*\frac{3}{4}* \dots *\frac{99}{100}\right)\left(\frac{2}{3}*\frac{4}{5}* \dots \frac{100}{101}\right)=\frac{1}{101} \Rightarrow A<\frac{1}{10}$

But I don't know, how to prove the other side?

Answer 1

$$A^2=\prod_{k=1}^{50}\frac{(2k-1)^2}{(2k)^2}=\frac{1}{200}\prod_{k=2}^{50}\frac{(2k-1)^2}{(2k-2)\cdot(2k)}>\frac{1}{200}\prod_{k=2}^{50}\frac{(2k-1)^2}{(2k-1)^2}=\frac{1}{200}>\frac{1}{225}$$