show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

Solution 1:

Note that $$\frac{n-1}{n}\gt\frac{n-2}{n-1}$$ for $n\gt 1$. So, we have $$\begin{align}\left(\frac 12\times \frac 34\times \cdots\times \frac{99}{100}\right)^2&\gt \left(\frac 12\times \frac 34\times\frac 56\times \cdots\times \frac{99}{100}\right)\left(\color{red}{\frac 12}\times \frac 23\times \frac{4}{5}\times\cdots\times \frac{98}{99}\right)\\&=\frac{1}{200}\\&\gt\frac{1}{225}\\&=\frac{1}{15^2}\end{align}$$

Solution 2:

Don't you already have the answer?

$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2>\frac{1}{2}\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{98}{99}\right)=\frac{1}{200}$