sigma k=2 to infinity, 1/k^2-1

how do you break up the k^2-1 into a minus part? i cant do it theres no partial fraction form i can ffind? please help with a little bit (Rofl)

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- Jan 5th 2011, 04:00 AMmathcoretelescopic series
sigma k=2 to infinity, 1/k^2-1

how do you break up the k^2-1 into a minus part? i cant do it theres no partial fraction form i can ffind? please help with a little bit (Rofl) - Jan 5th 2011, 04:18 AMtonio
- Jan 5th 2011, 04:24 AMchisigma
Is...

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

... so that...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 5th 2011, 04:33 AMmathcore
how do you work out this part? how would you make that jump because i'm fine with the difference of two squares part but i couldnt reach the next part myself. the reason i ask how is because i dont wanna have to ask for the help for each question if i know the logic i can do my remaining questions just fine

thank you :) - Jan 5th 2011, 04:47 AMmathcore
also is 1/2 sigma (then the sum expression) the same as sigma 1/2 (then the sum expression). we were instructed to put the 1/2 to the left of the sigma, is this ok

- Jan 5th 2011, 04:58 AMLiverpool
$\displaystyle \frac{1}{(k-1)(k+1)}=\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)$ using Partial Fractions Expansion.

and if c is a constant, then $\displaystyle \sum c \, a_k=c \, \sum a_k$. - Jan 5th 2011, 05:00 AMmathcore
im stuck on another one:

it is sigma k=2 to infinity again, but the sum is 1/k^3-k

the reason i am stuck is the partial fracs give three parts then i dont know what to do

i solved it as -1/k + (1/2)/k+1 + (1/2)/k-1, i am lost because that doesnt really help me at all - Jan 5th 2011, 05:01 AMMiss
Do not put new problem in the same thread.

Post a new thread for the new one. - Jan 5th 2011, 05:06 AMmathcore
- Jan 5th 2011, 05:10 AMMiss
This is the rules of this forum.

If you post new problems in the same thread, It will be too hard to follow the replies.

Anyway, as a student taking infinite series, you should be able to find the partial fraction expansion.

$\displaystyle \dfrac{1}{k^3-k}=\dfrac{1}{k(k^2-1)}=\dfrac{1}{k(k-1)(k+1)}=\dfrac{A}{k}+\dfrac{B}{k-1}+\dfrac{C}{k+1}$

You do not know how to find A,B & C ? - Jan 5th 2011, 05:48 AMmathcore
- Jan 5th 2011, 05:50 AMmathcore
what you have done there is exactly what i did but that form does not cancel out

- Jan 5th 2011, 05:52 AMmathcore
also i am not "taking infinite series", i did not say that. i did not even say i was a student. do not make these assumptions.

these are isolated problems that have not had any teaching hours dedicated to them - Jan 5th 2011, 05:55 AMGeneral
What is the statement of the problem?

You want to find its sum or just test its convergence? - Jan 5th 2011, 06:36 AMmathcore
find its sum

it is stated to be telescopic and as soon as i can see how the terms cancel out, i can find the sum, thats the easy part. but right now i cant see how to get them in a cancelling form