# Math Help - Evaluation of an Infinite Sum?

1. ## 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.

2. Originally Posted by TMFKAN64
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 $f(x+2\pi)=f(x)$.

3. 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.)

4. Originally Posted by TMFKAN64
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.