Therth term of a series (forr=1, 2, ...) is $\displaystyle \frac{4}{4r^2-1}$. Prove that the sumSof the firstnterms of the series is given by

$\displaystyle S=\frac{4n}{2n+1}$

Show that $\displaystyle 0<\frac{4}{4r^2-1}-\frac{1}{r^2}\leq\frac{1}{4r^2-1}$

Hence prove that the error in the approximate formula

$\displaystyle 1^{-2}+2^{-2}+...+n^{-2}\simeq S$

I have foundScorrectly, my proof of the second part is not quite satisfactory to me, and I don't know how to do the last part.

Proof:

$\displaystyle \frac{4}{4r^2-1}-\frac{1}{r^2}=\frac{1}{r^2(4r^2-1)}$

For allr, $\displaystyle r^2$ is positive, hence $\displaystyle \frac{1}{r^2(4r^2-1)}$is positive.

When $\displaystyle 0<r<\infty$, $\displaystyle \frac{1}{r^2(4r^2-1)}>0$, hence $\displaystyle 0<\frac{4}{4r^2-1}-\frac{1}{r^2}$

For allr, $\displaystyle \frac{1}{r^2(4r^2-1)}\leq\frac{1}{4r^2-1}$

Hence, $\displaystyle 0<\frac{4}{4r^2-1}-\frac{1}{r^2}\leq\frac{1}{4r^2-1}$.

For this last part, I did find the general term, but don't know how to find the sum of the series.

$\displaystyle \sum^n_{r=1} (\frac{1}{r})^2\simeq S$

Thanks!