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Summing series using contour integrals

  1. Dec 9, 2005 #1
    I would be grateful if someone can help me take this sum.

    Evaluate the following sum

    1/(n^2-4)^2, the sum goes over all positive integers n= 1,3,4,5...infty except 2.

    If it weren't squared I think I can figure it out( I remember my professor telling something about mutliplying by a factor 1/(e^z-1) ). How do I set the problem up? . I 'd appreciate any help you can give.

    I have split LHS into partial fractions and I am getting 4terms. It is of the form

    A/(n-2)^2 + B/(n+2)^2 +C/(n-2) + D/(n+2). And I know how sum series of type 1/n^2+a^2 using resdiue calculus. Please help.

  2. jcsd
  3. Dec 10, 2005 #2


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    Combine the C and D terms, you'll get a sum like 1/(n^2-a^2), can you handle this using residues? You can evaluate the A and B terms using the usual 1/n^2 sum (which can also be evaluated using residues).

    However, you can go at 1/(n^2-4)^2 directly. What would you use to sum something like 1/(n^2+a^2) and why don't you think it will work directly on 1/(n^2-4)^2?
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