Balti Tee 2020 · Ülesanne 1

Let $b_n = \frac{1}{a_n^2}$. Then $b_0 = \frac{1}{a_0^2}$ and

$$b_n = \frac{1 + 2020 \cdot a_{n-1}^2}{a_{n-1}^2} = b_{n-1} \left( 1 + 2020 \cdot \frac{1}{b_{n-1}} \right) = b_{n-1} + 2020.$$

Hence $b_{2020} = b_0 + 2020^2 = \frac{1}{a_0^2} + 2020^2$ and $a_{2020}^2 = \frac{1}{\frac{1}{a_0^2} + 2020^2} < \frac{1}{2020^2}$ which shows that $a_{2020} < \frac{1}{2020}$.