Bound on the tail of sum of inverse squares

Question

I am looking to bound the term $$\sum_{j\geq Cs}^N \frac{s^2}{j^2}$$ where $C\geq 1$ and $s$ is a positive real number. Obviously, we have $$\sum_{j\geq Cs}^N \frac{s^2}{j^2} \leq s^2\sum_{j\geq Cs}^N\frac{1}{j^2} \leq \frac{\pi^2}{6}s^2.$$ However, I wonder if it would be possible to obtain a lower order bound of the form $ks$, where $k$ is some constant that is independent of $N$.

Answer 1

$$\sum_{j \geq Cs}{\frac{s^2}{j^2}} \leq s^2\sum_{j \geq Cs}{\frac{1}{j(j-1)}}= \frac{s^2}{Cs-1}.$$