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Looking for the limit if it exists

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex] \sum_{n=2}^{\infty} \frac{2}{n^2-1}[/tex]

    3. The attempt at a solution

    I don't know what it is about this, maybe just because its late, but I can't figure this out. Can someone please help. Just looking for the limit if it exists. I have tried to figure it out; just not sure what to do. I wrote down the the terms of the sequence and the sum, but I can't see a pattern.
    Last edited: Nov 13, 2008
  2. jcsd
  3. Nov 13, 2008 #2
    Re: Series

    Note that: [tex] \sum_{k=1}^{\infty} \frac{1}{k^2}= \frac{ \pi ^2}{6} [/tex]
  4. Nov 13, 2008 #3
    Re: Series

    Note the identity [tex] \cot x = \sum_{n=-\infty}^{\infty} \frac{1}{x+n \pi} = \frac{1}{x} + \sum_{n=1}^{\infty} \frac{2x}{x^2 - n^2 \pi^2}[/tex], which can be proved by differentiating ln(sin(x)) using the infinite product form of sin(x).

    Then, using coth(x) = i cot(ix) we get [tex] \coth x = \frac{1}{x} + \sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2 \pi^2}[/tex], which may be useful for your problem.
    Last edited: Nov 13, 2008
  5. Nov 13, 2008 #4
    Re: Series

    Oops.. you can use the integral test if you just want to check its convergence.
    Last edited: Nov 13, 2008
  6. Nov 13, 2008 #5
    Re: Series

    Thanks I figured this out today though, and just for completeness I will post the way I solved it.

    If you use partial fraction decomposition to split of the initial fraction it becomes a telescoping series.

    [tex]\frac{2}{n^2-1} = \frac{2}{(n+1)(n-1)}[/tex]
    [tex]\frac{2}{(n+1)(n-1)} = \frac{A}{n+1} + \frac{B}{n-1}[/tex]

    [tex]2 = A(n-1) + B(n+1)[/tex]

    Solve that for A and B and you get A = -1, B=1.

    So the original sum becomes:

    [tex]\sum_{n=2}^{\infty} \frac{1}{n-1} - \frac{1}{n+1}[/tex]

    if you write plug in some terms and write out the series you will get:

    [tex]S_{n} = 1 + \frac{1}{2} -\frac{1}{n} -\frac{1}{n+1}[/tex]

    [tex]\lim_{x -> \infty} S_{n} = 1 + \frac{1}{2} = 3/2[/tex]
  7. Nov 13, 2008 #6
    Re: Series


    I thought of something like:

    Sn= 1-1/2+1/2-1/3 +1/3-1/4 +1/4-1/5+...=1

    \\edit: your sum in the first post is different than in the fifth!
  8. Nov 13, 2008 #7


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    Gold Member

    Re: Series

    Except that's not the telescoping sum that he found. So you would get

    1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6

    Confusion abounds however, is the series in the original post not the one you were trying to sum?
  9. Nov 13, 2008 #8
    Re: Series

    Yes, you're right I didn't read well enough!
  10. Nov 13, 2008 #9
    Re: Series

    Oh yeah sorry, in the very first post it should be a minus sign. Boy I guess that really makes a big difference.
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