Is it true that $\sum_{j=n}^{\infty}1/j^2=O(1/n)$?

Question

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

Answer 1

@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.

Answer 2

Take a look at this inequality for the integral comparison test. It yields

$$\frac{1}{n} \le \sum \limits_{j = n}^\infty \frac{1}{j^2} \le \frac{1}{n^2} + \frac{1}{n}$$

Answer 3

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$\ds{\sum_{j = n}^{\infty}{1 \over j^{2}} = \,\mrm{O}\pars{1 \over n}:\ {\large ?}}$.

Note that ( with Stolz-Ces$\grave{a}$ro Theorem ) \begin{align} \lim_{n \to \infty}\pars{n\sum_{j = n}^{\infty}{1 \over j^{2}}} & = \lim_{n \to \infty}{\sum_{j = n}^{\infty}1/j^{2} \over 1/n} = \lim_{n \to \infty}{\sum_{j = n + 1}^{\infty}1/j^{2} - \sum_{j = n}^{\infty}1/j^{2} \over 1/\pars{n + 1} - 1/n} \\[5mm] & = \lim_{n \to \infty}\braces{-1/n^{2} \over -1/\bracks{n\pars{n + 1}}} = \lim_{n \to \infty}\pars{1 + {1 \over n}} = \bbx{\ds{1}} \end{align}