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Math Help - Evaluation of an Infinite Sum?

  1. #1
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    Evaluation of an Infinite Sum?

    I'm looking to evaluate the infinite sum of 1/(n-f)^2 where n is the index of summation from 1 to infinity and f is a constant between zero and one.

    I really have no clue where to start on this. I've seen the summation evaluated to pi^2/6 if f=0 by using a Fourier transform of x^2, but I can't see how to adapt this for an arbitrary constant.

    Any ideas or hints would be appreciated.
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  2. #2
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    Quote Originally Posted by TMFKAN64 View Post
    I'm looking to evaluate the infinite sum of 1/(n-f)^2 where n is the index of summation from 1 to infinity and f is a constant between zero and one.

    I really have no clue where to start on this. I've seen the summation evaluated to pi^2/6 if f=0 by using a Fourier transform of x^2, but I can't see how to adapt this for an arbitrary constant.

    Any ideas or hints would be appreciated.
    Maybe, ... it would seem reasonable to find the Fourier series for the function f(x)=(x-f)^2 defined on [-\pi,\pi] and with periodic property  f(x+2\pi)=f(x).
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    That was my initial thought, but it doesn't really help. In the Fourier transform method when f=0, we end up finding that x^2 = pi^2/3 + 4 sum (-1)^n/n^2 cos nx. This becomes our sum when we let x=pi. If I try (x-f)^2 instead, I end up with some sin terms and a different constant offset, but the denominator is still n^2. (I get (x-f)^2 = pi^2/3 + f + 4 sum [(-1)^n/n^2 cos nx + f pi/n sin nx as the fourier series.)
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  4. #4
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    Quote Originally Posted by TMFKAN64 View Post
    That was my initial thought, but it doesn't really help. In the Fourier transform method when f=0, we end up finding that x^2 = pi^2/3 + 4 sum (-1)^n/n^2 cos nx. This becomes our sum when we let x=pi. If I try (x-f)^2 instead, I end up with some sin terms and a different constant offset, but the denominator is still n^2. (I get (x-f)^2 = pi^2/3 + f + 4 sum [(-1)^n/n^2 cos nx + f pi/n sin nx as the fourier series.)
    First it is more correct to say,
    \sum_{n=1}^{\infty} \frac{1}{(n-f)^2}
    Is undefined if f is an integer.
    So we should use a plus sign and state that f is a positive integers or non-integral real number.
    Note that if f is an positive integer then you get the standard "Basel problem" only shifted over.
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