By AM-GM $\frac {1}{i^2 + 2020} \leq \frac{2}{(i + \sqrt{2020})^2}$ and also $\frac{2}{ ( i + \sqrt{2020} )^2} < 2 \big ( \frac{1}{ i + \sqrt{2020} -1} - \frac{1}{i + \sqrt{2020} }\big )$ so we have $$ \sum \frac{1}{i^2 + 2020} < 2 ( \sum (\frac {1}{i + \sqrt{2020} -1} - \frac {1}{i + \sqrt{2020} } ) < 2 \times \frac{1}{1 + \sqrt{2020} -1} < \frac{1}{22}$$