Show that for any natural number the sum

$\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots + \frac{1}{n^2}$

lies between the values

$\displaystyle \left(1-\frac{2}{n+1}\right)\left(1-\frac{2}{2n+1}\right)\frac{\pi^2}{6}$ and $\displaystyle \left(1-\frac{1}{2n+1}\right)\left(1+\frac{1}{2n+1}\right) \frac{\pi^2}{6}.$