Prove that $\frac{1}{2020} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2019}{2020} < \frac{1}{44}$

Question

Prove that $\frac{1}{2020} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2019}{2020} < \frac{1}{44}$

I have proven the first half of the inequality, which is the expression is less than $\frac{1}{2020}$ by using the following method:

$\frac{1}{2} \times \frac{3}{4} \times \cdot \times \frac{2017}{2018} \frac{2019}{2020}$ and $\frac{2}{3} \times \frac{4}{5} \times \cdot \times \frac{2018}{2019}$ are both less than $1$, so their product = $\frac{1}{2019} \times \frac{2019}{2020}$ = $\frac{1}{2020}$ is smaller than both of them. Therefore $\frac{1}{2} \times \frac{3}{4} \times \cdot \times \frac{2017}{2018} \frac{2019}{2020}>\frac{1}{2020}$

However, I am unable to use the same idea to prove the other half of the inequality. Any ideas? Thank you in advance.

Answer 1

Because $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2019}{2020}=\sqrt{\frac{1\cdot3}{2^2}\cdot\frac{3\cdot5}{4^2}\cdot\frac{5\cdot7}{6^2}\cdot...\cdot\frac{2017\cdot2019}{2018^2}\cdot\frac{2019}{2020^2}}<$$ $$<\sqrt{\frac{2019}{2020^2}}<\frac{1}{\sqrt{2020}}<\frac{1}{44}.$$