Prove that there exists a constant $c>0$ such that the following happens

Question

Prove that there exists a constant $c>0$ such that for all $x \in [1,\infty)$, $$\sum_{n>x} \frac{1}{n^{2}} \leq \frac{c}{x}$$

Answer 1

Note that since $1/n^2$ is monotonically decreasing, we can assert that for $x\ge 2$

$$\sum_{n= x}^\infty \frac1{n^2}\le \int_{x-1}^\infty \frac{1}{t^2}\,dt=\frac1{x-1}\le \frac2x \tag 1$$

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For $x=1$, we have

$$\sum_{n=1}^\infty \frac1{n^2}\le 1+\int_{1}^\infty \frac{1}{t^2}\,dt=2 \tag 2$$

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Putting together $(1)$ and $(2)$, we have for all $x\ge 1$

$$\sum_{n= x}^\infty \frac1{n^2}\le\frac2x $$

as was to be shown!

Answer 2

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With the identity $\ds{\left.\sum_{n = 1}^{N}{1 \over n^{s}} \right\vert_{ {\large{\Re\pars{s}\ >\ 0 \atop N = 1,2,3,\ldots}}} = {N^{1 - s} \over 1 - s} + \zeta\pars{s} + s\int_{N}^{\infty}{\braces{x} \over x^{s + 1}}\,\dd x}$:

\begin{align} \sum_{n = m + 1}^{\infty}{1 \over n^{2}} & = \zeta\pars{2} - \sum_{n = 1}^{m}{1 \over n^{2}} = \zeta\pars{2} - \bracks{-\,{1 \over m} + \zeta\pars{2} + 2\int_{m}^{\infty}{\braces{x} \over x^{3}}\,\dd x} \\[5mm] & = {1 \over m}\ -\ \underbrace{2\int_{m}^{\infty}{\braces{x} \over x^{3}}\,\dd x} _{\ds{>\ 0\ \mbox{and}\ <\ {1 \over m^{2}}}}\implies \bbx{\ds{% {1 \over m} - {1 \over m^{2}} < \sum_{n = m + 1}^{\infty}{1 \over n^{2}} < \color{#f00}{{1 \over m}}}} \end{align}