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.

Since the question includes a hint that the goal is to study the distribution of square-free integers, here is an observation.
The Euler product for the Dirichlet series $$L(s) = \sum_{n\ge 1} \frac{q_n}{n^s}$$ with $q_n$ the indicator function of positive integers that are square-free is given by
$$L(s) = \prod_p \left(1+\frac{1}{p^s}\right)
= \prod_p \frac{1-1/p^{2s}}{1-1/p^s} = \frac{\zeta(s)}{\zeta(2s)}.$$
It follows by the Wiener-Ikehara theorem that 
$$\sum_{k=1}^{N} q_k \sim \mathrm{Res}(L(s); s=1) \frac{N^1}{1}
= \frac{1}{\zeta(2)} N = \frac{6}{\pi^2} N.$$
