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.

The tail is 
$$\frac{1}{N^2}+\frac{1}{(N+1)^2}+\frac{1}{(N+2)^2}+\cdots.$$
For $N\gt 1$ this is less than 
$$\frac{1}{(N-1)(N)}+ \frac{1}{(N)(N+1)}+ \frac{1}{(N+1)(N+2)}+\cdots.$$
Note that $\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$. Using this we find that the tail is less than
$$\left(\frac{1}{N-1}-\frac{1}{N}\right)+ \left(\frac{1}{N}-\frac{1}{N+1}\right)+ \left(\frac{1}{N+1}-\frac{1}{N+2}\right)+\cdots.$$
Open up the brackets and note the massive cancellation (telescoping). We find that the tail is less than $\frac{1}{N-1}$. 
Another way: Draw a picture of the curve $y=\frac{1}{x^2}$. Note that the area of the rectangle of height $\frac{1}{n^2}$ that goes from $x=n-1$ to $x=n$ is $\frac{1}{n^2}$. 
This rectangle is entirely below the curve on the interval from $n-1$ to $n$. Thus
the tail is less than
$$\int_{N-1}^\infty \frac{dx}{x^2}.$$
Evaluate the integral. We get $\frac{1}{N-1}$.  
