telescopic series

**mathcore** · January 5th 2011, 04:00 AM

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

**tonio** · January 5th 2011, 04:18 AM

Originally Posted by mathcore

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

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

Tonio

**chisigma** · January 5th 2011, 04:24 AM

Is...

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

... so that...

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

Kind regards

$\chi$ $\sigma$

**mathcore** · January 5th 2011, 04:33 AM

Originally Posted by tonio

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

Tonio

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

**mathcore** · January 5th 2011, 04:47 AM

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

**Liverpool** · January 5th 2011, 04:58 AM

$\frac{1}{(k-1)(k+1)}=\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)$ using Partial Fractions Expansion.

and if c is a constant, then $\sum c \, a_k=c \, \sum a_k$ .

**mathcore** · January 5th 2011, 05:00 AM

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

**Miss** · January 5th 2011, 05:01 AM

Do not put new problem in the same thread.
Post a new thread for the new one.

**mathcore** · January 5th 2011, 05:06 AM

Originally Posted by Miss

Do not put new problem in the same thread.
Post a new thread for the new one.

its the same question just the next one

**Miss** · January 5th 2011, 05:10 AM

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.

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

You do not know how to find A,B & C ?

**mathcore** · January 5th 2011, 05:48 AM

Originally Posted by Miss

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.

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

You do not know how to find A,B & C ?

please do not tell me what i should or shouldnt be able to find, that is not yours to say.

yes i can do partial fractions but i cant seem to arranging into a telescopic form

**mathcore** · January 5th 2011, 05:50 AM

what you have done there is exactly what i did but that form does not cancel out

**mathcore** · January 5th 2011, 05:52 AM

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

**General** · January 5th 2011, 05:55 AM

What is the statement of the problem?
You want to find its sum or just test its convergence?

**mathcore** · January 5th 2011, 06:36 AM

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

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