Is it true that $\sum_{j=n}^{\infty}1/j^2=O(1/n)$? I've been trying to prove or disprove that $$\sum_{j=n}^{\infty}\frac{1}{j^2}=O\bigg(\frac{1}{n}\bigg),$$
but have failed. Actually, what is
$$\lim_{n\to \infty}n\sum_{j=n}^{\infty}\frac{1}{j^2}?$$

Take a look at this inequality for the integral comparison test. It yields
$$\frac{1}{n} \le \sum \limits_{j = n}^\infty \frac{1}{j^2} \le \frac{1}{n^2} + \frac{1}{n}$$
