Trying to get a bound on the tail of the series for $\zeta(2)$ $$\frac{\pi^2}{6} = \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2}$$ I hope we agree.
Now how do I get a grip on the tail end $\sum_{k \geq N} \frac{1}{k^2}$ which is the tail end which goes to zero?
I want to show that $\sqrt{x}\cdot \mathrm{tailend}$ is bounded as $x \to \infty$.
All this to show that $x\cdot \mathrm{tailend} = \mathcal O\sqrt{x}$
The purpose is to get the asymptotic formula for the distribution of square free integers.
p.269 Exercise 8 Stewart and Tall.

Maybe this should be a comment, but it is a bit better to use $$\frac{\pi^2}{12}=\eta(2)=\sum_{n\geqslant 1}(-1)^{n-1}\frac 1 {n^2}$$ and the usual bound for alternating series in this case. 
