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I am stuck on how to calculate the value of the following sum:

$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$

I am aware that you need to find the corresponding function whose Fourier series is represented by the above, and then use the Parseval Identity to find the actual value similar to how you show that $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$.

I believe finding the corresponding function has something to do with how the Fourier series is a convolution of the function and the Dirichlet kernel, but beyond that I'm stuck. Any hints are appreciated.

Edit: We need to consider the Fourier series of a function where for two values a1 and a2, f(x) = a1 if x $\geq$ 0 and f(x) = a2 if x < 0.

Newbie1000
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  • Relate $\sum_{n\ge 0} \frac{e^{i(2n+1)x}}{2n+1}$ to the power series of $-\log(1-z)$ (or $1/(1-z)$ after a differentiation) – reuns Jan 03 '23 at 21:14
  • I edited my question because I left out some important information. – Newbie1000 Jan 03 '23 at 21:19
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    Note the Taylor series $\displaystyle \arctan x = \sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}$ for $|x|\le1$. So your sum is $\arctan(1)=\frac{\pi}{4}$. – Abezhiko Jan 03 '23 at 21:24
  • @Abezhiko Thank you this really helps me understand it. To that end, is there a theorem that somehow relates the Taylor Series to the Fourier Series? – Newbie1000 Jan 03 '23 at 21:41
  • The x term in the series expansion you provided must relate to the Dirichlet kernel, right? – Newbie1000 Jan 03 '23 at 21:42
  • @Newbie1000 I think you can find useful information here: https://math.stackexchange.com/questions/7301/connection-between-fourier-transform-and-taylor-series – Henry Jan 03 '23 at 22:13
  • @Newbie1000 I'm not aware of a precise formula, but somehow it would need to expand the polynomial basis in the Fourier basis, something like $x^n = \displaystyle \sum_{k\in\mathbb{Z}} e^{ikx} \langle e^{ikx},x^n \rangle$; I mean, if you have the Fourier series of powers of $x$, you're basically done. – Abezhiko Jan 04 '23 at 07:09

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Hint: expand in fourier series the absolute value of x, $f(x)=|x|$ in the interval $\pi \leq x \leq \pi$

Marc Navarro
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