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Math Help - [SOLVED] Series Question

  1. #1
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    [SOLVED] Series Question

    I'm trying to find the sum of this infinite series (with n=2) of
    6/(n^2-1)

    for some reason my answer isn't matching up with the answer choices.

    What I did:
    1/(n+1)-1/(n-1)
    Sn=(1/3-1)+(1/4-1/2)+(1/5 - 1/3)+.....+(1/(n-1)-1/(n-3))+(1/n - 1/(n-2))+(1/(n+1)- 1/(n-1))

    after canceling things out and taking the limit I ended up with -1-(1/2)=-3/2

    Can somebody please tell me what I'm doing wrong. Any help is appreciated, thank you!
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  2. #2
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    Quote Originally Posted by yzobel View Post
    I'm trying to find the sum of this infinite series (with n=2) of
    6/(n^2-1)

    for some reason my answer isn't matching up with the answer choices.

    What I did:
    1/(n+1)-1/(n-1)
    Sn=(1/3-1)+(1/4-1/2)+(1/5 - 1/3)+.....+(1/(n-1)-1/(n-3))+(1/n - 1/(n-2))+(1/(n+1)- 1/(n-1))

    after canceling things out and taking the limit I ended up with -1-(1/2)=-3/2

    Can somebody please tell me what I'm doing wrong. Any help is appreciated, thank you!
    The correct expression 3 \left( \frac{1}{n - 1} - \frac{1}{n+1}\right). Now, if you do the arithmetic correctly you should get the correct answer.
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