If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}
If $$x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$$
show that
$$\frac{201}{403}<x<\frac{2014}{2015}$$
So, I manage to do the RH inequality using that $$x<\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2014\cdot2015}=1-\frac{1}{2015}=\frac{2014}{2015}$$ Unfortunately I can't find a way to do the LH inequality. I saw that
$$\frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ for 1005 times is equal to $\frac{201}{403}$ but I don't see why
$$x> \frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ thx!

We have
$$\frac{1}{k}-\frac{1}{k+1} < \frac{1}{k^2} < \frac{1}{k-1} - \frac{1}{k}$$
and so (telescoping sum)
$$\frac{1}{2} - \frac{1}{2016} < \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{1}{1} - \frac{1}{2015}$$
that is
$$\frac{1007}{2016}< \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{2014}{2015}$$
Now,  LHS $= \frac{2014}{4032}= \frac{201.4}{403.2}> \frac{201}{403}$ (mediant)
