Prove that $\exists c>0$ s.t $\sum_{n\geq x}\frac{1}{n^2}\leq \frac{c}{x}$ $\mathbf{Question:}$ Prove that there exists a constant $c>0$ such that for all $x \in [1, \infty)$, $\displaystyle\sum_{n \geq x}\frac{1}{n^2} \leq \frac{c}{x}$
$\mathbf{Attempt:}$ Evidently, 
$\displaystyle\bigg[\displaystyle\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]{\lfloor x \rfloor}\leq x\displaystyle\sum_{n \geq x}\frac{1}{n^2} \leq \displaystyle\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]{\lceil x \rceil} $. 
This inequality arises since, 
$\displaystyle\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]=\displaystyle\sum_{n \geq x}\frac{1}{n^2}$.
Now, let $\lfloor x\rfloor=m$
$\displaystyle\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg]m$
$=\lim_{{m \to \infty},{r\to \infty}}\bigg[\frac{m}{(m+1)^2}+\frac{m}{(m+2)^2}+\frac{m}{(m+3)^2}+...+\frac{m}{(m+r)^2}+...\bigg]=$
$\lim_{{h \to 0},{r\to \infty}}h\bigg[\frac{1}{(1+h)^2}+\frac{1}{(1+2h)^2}+\frac{1}{(1+3h)^2}+...+\frac{1}{(1+rh)^2}+...\bigg]=\displaystyle\int_{x=0}^\infty\frac{1}{(1+x)^2}dx=1$
We have, $\lceil x\rceil=m+1$ and $\displaystyle\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg](m+1)=\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg]m$
Thereby, we conclude, $\displaystyle \lim_{x\to \infty}x\sum_{n \geq x}\frac{1}{n^2}=1 $.
Therefore, from the definition of limit, $\forall \varepsilon>0$, $\exists G>0$ such that $\bigg|x\displaystyle\sum_{n \geq x}\frac{1}{n^2}-1\bigg|<\varepsilon$ $\forall x>G$. Choosing $\varepsilon =1$, we get $G=G_0$ such that $x\displaystyle\sum_{n \geq x}\frac{1}{n^2}<2$, for any $x> G_0$. 
We set $\displaystyle c=\max\bigg\{2, \sup_{x\in [1,G_0]}x\sum_{n \geq x}\frac{1}{n^2} \bigg\}$
Is this procedure valid? Kindly verify. 
 A: 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.

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

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 $m=1$,
$$
\begin{align}
\sum_{n\ge m}\frac1{n^2}
&\le1+\sum_{n\ge 2}\frac1{(n-1)n}\\
&=1+\sum_{n\ge 2}\left(\frac1{n-1}-\frac1n\right)\\
&=2\\[6pt]
&=\frac2m
\end{align}
$$
A: I think you would be better off starting with an integral from the start:
$$\frac 1{n^2} \le \int_{n-1}^{n} \frac 1{t^2} \, dt.$$
For any integer $k$ [edited to address complaint] integer $k \ge 2$ you have
$$\sum_{n=k}^\infty \frac 1{n^2} \le \int_{k-1}^\infty \frac 1{t^2} \, dt = \frac 1{k-1}$$
From here it is not too difficult.
