Prove that there exists a constant $c>0$ such that the following happens
Note that since $1/n^2$ is monotonically decreasing, we can assert that for $x\ge 2$
$$\sum_{n= x}^\infty \frac1{n^2}\le \int_{x-1}^\infty \frac{1}{t^2}\,dt=\frac1{x-1}\le \frac2x \tag 1$$
For $x=1$, we have
$$\sum_{n=1}^\infty \frac1{n^2}\le 1+\int_{1}^\infty \frac{1}{t^2}\,dt=2 \tag 2$$
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!