1. Telescopic sum help needed

Need some slight help on the following sum, this is my first post so i hope ive put this in the correct place

sum of from k=2 to infinity (1/(k^3-k)

currently i have the following, ive turned it to a partial fraction and obtained
sum of from k=2 to infinity (k/(k^2-1)-(1/k)
Then from this entering numerical values for k:
(2/3-1/2+3/8-1/3......) and so on however cant see any relation between numbers cancelling etc like in previous questions i have done.

Thanks in advance for any help.

2. Originally Posted by breitling
Need some slight help on the following sum, this is my first post so i hope ive put this in the correct place

sum of from k=2 to infinity (1/(k^3-k)

currently i have the following, ive turned it to a partial fraction and obtained
sum of from k=2 to infinity (k/(k^2-1)-(1/k)
Then from this entering numerical values for k:
(2/3-1/2+3/8-1/3......) and so on however cant see any relation between numbers cancelling etc like in previous questions i have done.

Thanks in advance for any help.
Try splitting $\frac{k}{k^2-1}$ into partial fractions.

3. Try to prove:

$\displaystyle\Sigma_{k=2}^{n}\frac{1}{k^3-k}=\frac{n^2+n-2}{4n^2+4n}$

4. Originally Posted by Also sprach Zarathustra
Try to prove:

$\displaystyle\Sigma_{k=2}^{n}\frac{1}{k^3-k}=\frac{n^2+n-2}{4n^2+4n}$
Trying to go down both roots here, i think the one that im after is via the partial fraction but struggling. However going for the general rule and i obtained 1/(k/4(k(k^2+2k+1)-2(k+1)) not sure if this is correct from here to rearrange to get the formula stated. I used the sum of k up to infinity being k/2(k+1) then subtracted the sum up to 1, the same for k^3 using k^2/4(k+1)^2?

5. Originally Posted by alexmahone
Try splitting $\frac{k}{k^2-1}$ into partial fractions.
Think i may of got it via this actually, use doing the calculation....

6. Originally Posted by alexmahone
Try splitting $\frac{k}{k^2-1}$ into partial fractions.
i split that up into partial fractions getting the sum of from 2 to infinity 1/2(k-1)-1/2(k+1) which turns out i think to be 1/2(1+1/2)=3/4 then we have the subtraction of k=2 to infinity for 1/k, i decided to put it as one sum and try to spot of patterns etc as i have done previous hence sum of k=2 to infinity of
1/2( 1/(k-1)-1/(k+1)-2/k) to get the series of 1/2(1-1/3-1+1/2-1/4/-2/3+1/3-1/5-1/2....) a lot of the terms cancel but cant see anything of such so not sure what to do....

7. Originally Posted by breitling
i split that up into partial fractions getting the sum of from 2 to infinity 1/2(k-1)-1/2(k+1) which turns out i think to be 1/2(1+1/2)=3/4 then we have the subtraction of k=2 to infinity for 1/k, i decided to put it as one sum and try to spot of patterns etc as i have done previous hence sum of k=2 to infinity of
1/2( 1/(k-1)-1/(k+1)-2/k) to get the series of 1/2(1-1/3-1+1/2-1/4/-2/3+1/3-1/5-1/2....) a lot of the terms cancel but cant see anything of such so not sure what to do....
$\frac{1}{k^3-k}=\frac{k}{k^2-1}-\frac{1}{k}$

$=\frac{k-1+1}{k^2-1}-\frac{1}{k}$

$=\frac{1}{k+1}+\frac{1}{k^2-1}-\frac{1}{k}$

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

8. Originally Posted by breitling
Need some slight help on the following sum, this is my first post so i hope ive put this in the correct place

sum of from k=2 to infinity (1/(k^3-k)

currently i have the following, ive turned it to a partial fraction and obtained
sum of from k=2 to infinity (k/(k^2-1)-(1/k)
Then from this entering numerical values for k:
(2/3-1/2+3/8-1/3......) and so on however cant see any relation between numbers cancelling etc like in previous questions i have done.

Thanks in advance for any help.
Is...

$\displaystyle \frac{1}{k^{3}-k} = \frac{1}{2}\ (\frac{1}{k-1} + \frac{1}{k+1}) - \frac{1}{k}$ (1)

... so that...

$\displaystyle \sum_{k=2}^{\infty} \frac{1}{k^{3}-k} = \frac{1}{2} + \frac{1}{6} - \frac{1}{2} + \frac{1}{4} + \frac{1}{8} - \frac{1}{3} + \frac{1}{6} + \frac{1}{10} - \frac{1}{4} +$

$\displaystyle + \frac{1}{8} + \frac{1}{12} - \frac{1}{5} + \frac{1}{10} + \frac{1}{14} - \frac{1}{6} + \frac{1}{12} + \frac{1}{16} - \frac{1}{7} + ... = \frac{1}{4}$ (2)

Kind regards

$\chi$ $\sigma$

9. Originally Posted by chisigma
Is...

$\displaystyle \frac{1}{k^{3}-k} = \frac{1}{2}\ (\frac{1}{k-1} + \frac{1}{k+1}) - \frac{1}{k}$ (1)

... so that...

$\displaystyle \sum_{k=2}^{\infty} \frac{1}{k^{3}-k} = \frac{1}{2} + \frac{1}{6} - \frac{1}{2} + \frac{1}{4} + \frac{1}{8} - \frac{1}{3} + \frac{1}{6} + \frac{1}{10} - \frac{1}{4} +$

$\displaystyle + \frac{1}{8} + \frac{1}{12} - \frac{1}{5} + \frac{1}{10} + \frac{1}{14} - \frac{1}{6} + \frac{1}{12} + \frac{1}{16} - \frac{1}{7} + ... = \frac{1}{4}$ (2)

Kind regards

$\chi$ $\sigma$
I got this also, thanks for the help guys.