# Thread: Error in approximate formula

1. ## Error in approximate formula

The rth term of a series (for r=1, 2, ...) is $\frac{4}{4r^2-1}$. Prove that the sum S of the first n terms of the series is given by
$S=\frac{4n}{2n+1}$
Show that $0<\frac{4}{4r^2-1}-\frac{1}{r^2}\leq\frac{1}{4r^2-1}$
Hence prove that the error in the approximate formula
$1^{-2}+2^{-2}+...+n^{-2}\simeq S$

I have found S correctly, my proof of the second part is not quite satisfactory to me, and I don't know how to do the last part.
Proof:
$\frac{4}{4r^2-1}-\frac{1}{r^2}=\frac{1}{r^2(4r^2-1)}$
For all r, $r^2$ is positive, hence $\frac{1}{r^2(4r^2-1)}$is positive.
When $0, $\frac{1}{r^2(4r^2-1)}>0$, hence $0<\frac{4}{4r^2-1}-\frac{1}{r^2}$
For all r, $\frac{1}{r^2(4r^2-1)}\leq\frac{1}{4r^2-1}$
Hence, $0<\frac{4}{4r^2-1}-\frac{1}{r^2}\leq\frac{1}{4r^2-1}$.

For this last part, I did find the general term, but don't know how to find the sum of the series.
$\sum^n_{r=1} (\frac{1}{r})^2\simeq S$

Thanks!

2. Is...

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

$\displaystyle = \frac{1}{k-\frac{1}{2}} - \frac{1}{k+\frac{1}{2}}$ (1)

... so that is...

$\displaystyle \sum_{k=1}^{n} \frac{4}{4 k^{2} - 1} = \sum_{k=1}^{n} \frac{1}{k^{2} - \frac{1}{4}} = 2 - \frac{2}{3} + \frac{2}{3} - \frac{2}{5}+ ... - \frac{1}{n+\frac{1}{2}} = 2 (1-\frac{1}{2n+1}) =$

$\displaystyle \frac{4n}{2n+1}$ (2)

End of 'part 1'... to be continued...

Kind regards

$\chi$ $\sigma$

3. Now we have to find the 'error' ...

$\displaystyle e= \sum_{k=1}^{n} (\frac{1}{k^{2}-\frac{1}{4}} - \frac{1}{k^{2}})$ (1)

Is...

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

... so that is...

$\displaystyle e= \frac{1}{2}\ \{(1-\frac{1}{3}) + \frac{1}{4} (\frac{1}{3} - \frac{1}{5}) + \frac{1}{9} (\frac{1}{5} - \frac{1}{7}) + ... + \frac{1}{n^{2}} (\frac{1}{2n-1} - \frac{1}{2n+1}) \}$ (3)

At this point the question is: can we further 'simplify' the result (3)?...

Kind regards

$\chi$ $\sigma$

4. Thank you! it doesn't look very encouraging trying to simplify the result, but I'm supposed to prove that it is less than S/4. I thought it would have something to do with finding the sum of the series, which I don't know how, then finding the difference.

5. I think that the 'spirit' of the problem is the following. Let suppose we have 'forgotten' that is...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac {\pi^{2}}{6}$ (1)

... and we have to find some pratical method to compute the sum. The general term tend to 0 like $\frac{1}{n^{2}}$ and that means that we have to sum a very high number of terms. If however we consider that is...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} = \sum_{n=1}^{\infty} \frac{1}{n^{2}-\frac{1}{4}} - \sum_{n=1}^{\infty} \frac{1}{4\ n^{4}-n^{2}}$ (2)

... with the first series on the right which we know is equal to 2 and the second which has its general term that tends to 0 like $\frac{1}{n^{4}}$, our task becomes much less difficult ...

Kind regards

$\chi$ $\sigma$