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
$$\lim_{n\to \infty}n\sum_{j=n}^{\infty}\frac{1}{j^2}?$$

@Dominik's answer is fine, just another point of view, an elementary one.
One may observe that,
$$
\begin{align}
\frac{1}{j(j+1)}\leq \:&\frac{1}{j^2} \leq \frac{1}{j(j-1)} \qquad \qquad j=2,3,4,\cdots,
\\\\\frac{1}{j}-\frac{1}{j+1}\leq \:&\frac{1}{j^2} \leq \frac{1}{j-1}-\frac{1}{j} \qquad \qquad j=2,3,4,\cdots,
\end{align}
$$ by summing, one gets two telescoping sums obtaining
$$
\frac{1}{n}-\frac{1}{N+1}\leq \sum_{j=n}^N\frac{1}{j^2} \leq \frac{1}{n-1}-\frac{1}{N},\qquad N>1,n>1,
$$ letting $N \to \infty$ one obtains
$$
\frac{1}{n}\leq \sum_{j=n}^\infty\frac{1}{j^2} \leq \frac{1}{n-1},\qquad n>1,
$$ then, multiplying by $n$ and letting $n \to \infty$, one has
$$
\lim_{n \to \infty}n\sum_{j=n}^\infty\frac{1}{j^2}=1
$$ giving

$$
\sum_{j=n}^\infty\frac{1}{j^2}=\mathcal{O}\left( \frac1n\right), \qquad n \to \infty,
$$ 

as expected.
