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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. $$\sum_{n=2}^\infty{\frac4\pi}\frac{{\frac12}\cos(n\frac\pi2-nx)-{\frac12}\cos(n\frac\pi2+nx)}{1-n^2}$$ $$-\sum_{n=2}^\infty{\frac4\pi}\frac{\cos(n\frac\pi2-nx)}{2n^2-2}-\frac{\cos(n\frac\pi2+nx)}{2n^2-2}$$

Davor
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1 Answers1

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

robjohn
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    Mathematica agrees with this answer. – robjohn Mar 21 '19 at 20:01
  • 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 https://www.youtube.com/watch?v=bG6bpYlQx2s – Davor Mar 22 '19 at 09:13
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    @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? – robjohn Mar 22 '19 at 09:23
  • 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 – Davor Mar 22 '19 at 09:35
  • How did you get that the sum at the first line gives 1 for non integers? @robjohn – Lac Jan 16 '21 at 04:37
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    @LSS: I am not sure what you mean. Are you asking about this answer? – robjohn Jan 16 '21 at 07:06
  • $$ \begin{align} \sum_{n=1}^\infty\frac1{(4n)^2-1} &=\frac12\sum_{n=1}^\infty\left(\frac1{4n-1}-\frac1{4n+1}\right)\ &=\frac12\left[1+\sum_{n\in\mathbb{Z}}\frac1{4n-1}\right]\ \end{align} $$

    @robjohn i didn't understand how do you went from the first part, to the second :)

     – Lac Jan 16 '21 at 20:06
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    @LSS: $$ \begin{align} \sum_{n=1}^\infty\left(\frac1{4n-1}-\frac1{4n+1}\right) &=\lim_{m\to\infty}\sum_{n=1}^m\left(\frac1{4n-1}+\frac1{-4n-1}\right)\ &=\overbrace{\quad\ 1\quad\ \vphantom{\sum_m^m}}^{\substack{\text{this cancels}\\text{the $n=0$ term}}}+\lim_{m\to\infty}\overbrace{\sum_{n=-m}^m\frac1{4n-1}}^{\substack{\text{this sum includes}\\text{$n=0$, which is not}\\text{in the sum above}}}\ &=1+\sum_{n\in\mathbb{Z}}\frac1{4n-1} \end{align} $$ – robjohn Jan 16 '21 at 20:38