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
$$\lim_{n\to \infty}n\sum_{j=n}^{\infty}\frac{1}{j^2}?$$
 A: Take a look at this inequality for the integral comparison test. It yields
$$\frac{1}{n} \le \sum \limits_{j = n}^\infty \frac{1}{j^2} \le \frac{1}{n^2} + \frac{1}{n}$$
A: @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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\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}
