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Math Help - telescopic series

  1. #1
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    telescopic 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
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  2. #2
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    Quote Originally Posted by mathcore View Post
    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
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    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
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  4. #4
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    Quote Originally Posted by tonio View Post
    \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
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  5. #5
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    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
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  6. #6
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    \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.
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  7. #7
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    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
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  8. #8
    Member Miss's Avatar
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    Do not put new problem in the same thread.
    Post a new thread for the new one.
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  9. #9
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    Quote Originally Posted by Miss View Post
    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
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  10. #10
    Member Miss's Avatar
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    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 ?
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  11. #11
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    Quote Originally Posted by Miss View Post
    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
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  12. #12
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    what you have done there is exactly what i did but that form does not cancel out
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  13. #13
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    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
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  14. #14
    Super Member General's Avatar
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    What is the statement of the problem?
    You want to find its sum or just test its convergence?
    Last edited by General; January 5th 2011 at 06:18 AM.
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  15. #15
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    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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