Prove that $\exists c>0$ s.t $\sum_{n\geq x}\frac{1}{n^2}\leq \frac{c}{x}$

Comments

I had initially thought that, although $x$ is usually a real variable, $x$ was intended to be an integer. However, if $x$ is intended to take on any real value, one needs to be clear on exactly what is meant, since this is not a standard use for $\sum$. Here are two that come to mind at first: $$ \sum_{n\ge x}\frac1{n^2}=\sum_{n=\lceil x\rceil}^\infty\frac1{n^2} $$ and $$ \sum_{n\ge x}\frac1{n^2}=\sum_{n=0}^\infty\frac1{(n+x)^2} $$ That point aside, your approach of finding the limit and using that for all $x$ past a finite value, and then handling the initial finite range of $x$ separately, is quite valid.

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Slick Answer

For $m\ge1$, $$ \begin{align} \sum_{n\ge m}\frac1{n^2} &\le\sum_{n\ge m}\frac1{\left(n-\frac12\right)\left(n+\frac12\right)}\\ &=\sum_{n\ge m}\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\ &=\frac1{m-\frac12}\\[3pt] &\le\frac2m \end{align} $$

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Less Slick Answer

For $m\ge2$, $$ \begin{align} \sum_{n\ge m}\frac1{n^2} &\le\sum_{n\ge m}\frac1{(n-1)n}\\ &=\sum_{n\ge m}\left(\frac1{n-1}-\frac1n\right)\\ &=\frac1{m-1}\\[3pt] &\le\frac2m \end{align} $$ For $m=1$, $$ \begin{align} \sum_{n\ge m}\frac1{n^2} &\le1+\sum_{n\ge 2}\frac1{(n-1)n}\\ &=1+\sum_{n\ge 2}\left(\frac1{n-1}-\frac1n\right)\\ &=2\\[6pt] &=\frac2m \end{align} $$