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Given that $f(x) = x^2$, I got that the Fourier series is $$ x^2 = \dfrac{\pi^2}{3} + \sum\limits_{n=1}^{\infty} (-1)^n \cdot 4 \cdot \dfrac{\cos(nx)}{n^2} $$ for $x \in [-\pi, \pi]$. From the above series, how can I find $\sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)^2}$?

user466721
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1 Answers1

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Hint:

Consider $x = \pi$ and then $x = 0$. Then, add the two results and see what you get.

  • thank you... Yes I got that answer... Can You please tell me how to get the summation of 1/n^4 from the above series? – user466721 Jul 24 '17 at 09:31