I need to determine the sum $$\sum_{n=1}^\infty \left( \frac{1}{(4n)^2-1}-\frac{1}{(4n+2)^2+1}\right)$$ using the Fourier series of $\lvert \cos x\rvert$ on the interval $ [-\pi,\pi]$.

I have already calculated Fourier series and I get this: $$\lvert \cos x\rvert= {\frac2\pi}+\sum_{n=2}^\infty{\frac4\pi}\frac{\cos(n\frac\pi2)}{1-n^2}\cos(nx)$$

I do not know how to manipulate the Fourier series to get that specific sum. I tried this but did not get me anywhere. $$\sum_{n=2}^\infty{\frac4\pi}\frac{{\frac12}\cos(n\frac\pi2-nx)-{\frac12}\cos(n\frac\pi2+nx)}{1-n^2}$$ $$-\sum_{n=2}^\infty{\frac4\pi}\frac{\cos(n\frac\pi2-nx)}{2n^2-2}-\frac{\cos(n\frac\pi2+nx)}{2n^2-2}$$

  • $\begingroup$ $\cos(n \frac{\pi}{2}) = 1 \forall n \in \mathbb{N}$ $\endgroup$ – Max Mar 21 at 11:01
  • 2
    $\begingroup$ @Max This is by factor 4 far from truth. $\endgroup$ – user Mar 21 at 11:04
  • $\begingroup$ Sorry, lack of coffee. Was meant to be $=0$. $\endgroup$ – Max Mar 21 at 11:13
  • $\begingroup$ Its not 0, it can be 0, -1 or 1 depending on n $\endgroup$ – Davor Mar 21 at 11:20
  • $\begingroup$ Oh dammit, i'm getting coffee now! Sorry for the nonsense. $\endgroup$ – Max Mar 21 at 11:30

The sum of the first term is $$ \begin{align} \sum_{n=1}^\infty\frac1{(4n)^2-1} &=\frac12\sum_{n=1}^\infty\left(\frac1{4n-1}-\frac1{4n+1}\right)\\ &=\frac12\left[1+\sum_{n\in\mathbb{Z}}\frac1{4n-1}\right]\\ &=\frac12+\frac18\sum_{n\in\mathbb{Z}}\frac1{n-\frac14}\\ &=\frac12-\frac\pi8\cot\left(\frac\pi4\right)\\[6pt] &=\frac12-\frac\pi8\tag1 \end{align} $$ The sum of the second term is $$ \begin{align} \sum_{n=1}^\infty\frac1{(4n+2)^2+1} &=\frac i2\sum_{n=1}^\infty\left(\frac1{i-(4n+2)}+\frac1{i+(4n+2)}\right)\\ &=\frac i2\left[\sum_{n\in\mathbb{Z}}\frac1{i+(4n+2)}-\frac1{i+2}-\frac1{i-2}\right]\\ &=\frac i8\sum_{n\in\mathbb{Z}}\frac1{n+\frac{2+i}4}-\frac15\\ &=\frac{\pi i}8\cot\left(\pi\frac{2+i}4\right)-\frac15\\[3pt] &=\frac\pi8\tanh\left(\frac\pi4\right)-\frac15\tag2 \end{align} $$ where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).

Subtracting $(2)$ from $(1)$ yields $$ \sum_{n=1}^\infty\left(\frac1{(4n)^2-1}-\frac1{(4n+2)^2+1}\right)=\frac7{10}-\frac\pi8-\frac\pi8\tanh\left(\frac\pi4\right)\tag3 $$

  • $\begingroup$ Mathematica agrees with this answer. $\endgroup$ – robjohn Mar 21 at 20:01
  • $\begingroup$ Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s $\endgroup$ – Davor Mar 22 at 9:13
  • $\begingroup$ @Davor: you mean that the first sum should be $\sum\limits_{n=1}^\infty\frac1{16n^2-1}$? Are you going to alter the question, or ask another with the correct sum? $\endgroup$ – robjohn Mar 22 at 9:23
  • $\begingroup$ Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform $\endgroup$ – Davor Mar 22 at 9:35

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