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Deduce the interval of validity of the expansion:

$$\cos{x} = \frac 8 \pi \sum_{n=1}^{\infty} \frac {n}{4n^2 - 1} \sin(2nx)$$

I tried to apply some basic convergence tests but they don't seem to work.

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Note that $$\sum_{n\geq1}\frac{n}{4n^{2}-1}\sin\left(2nx\right)=\textrm{Im}\left(\sum_{n\geq1}\frac{ne^{2inx}}{4n^{2}-1}\right)$$ then we have to calculate the power series $$\sum_{n\geq1}\frac{nz^{2n}}{4n^{2}-1}=\frac{1}{4}\left(\sum_{n\geq1}\frac{2nz^{2n}}{2n-1}-\sum_{n\geq1}\frac{2nz^{2n}}{2n+1}\right)$$ $$=\frac{1}{4}\left(\sum_{n\geq1}\frac{z^{2n}}{2n-1}+\sum_{n\geq1}\frac{z^{2n}}{2n-1}\right)=\frac{z^{2}\tanh^{-1}\left(z\right)-z+\tanh^{-1}\left(z\right)}{4z},\,\left|z\right|\leq1$$ where $\tanh^{-1}\left(z\right)=\left[\log\left(1+z\right)-\log\left(1-z\right)\right]/2$ is the inverse hyperbolic tangent and the the logarithm function in this case is the complex logarithm with the principal value, then taking $z=e^{ix}$ $$\sum_{n\geq1}\frac{n}{4n^{2}-1}\sin\left(2nx\right)=\textrm{Im}\left(\frac{e^{2ix}\tanh^{-1}\left(e^{ix}\right)-e^{ix}+\tanh^{-1}\left(e^{ix}\right)}{4e^{ix}}\right)$$ $$=\textrm{Im}\left(\frac{2\cos\left(x\right)\tanh^{-1}\left(e^{ix}\right)-1}{4}\right)=\color{blue}{\frac{\cos\left(x\right)\left(\arg\left(1+e^{ix}\right)-\arg\left(1-e^{ix}\right)\right)}{4}}$$ so the problem boils down to solve the equation $$\arg\left(1+e^{ix}\right)-\arg\left(1-e^{ix}\right)=\frac{\pi}{2}.$$ We have, using the properties of the $\arg(z),$ $$\arg\left(1+e^{ix}\right)-\arg\left(1-e^{ix}\right)=\arg\left(i\cot\left(\frac{x}{2}\right)\right)$$ so if $\cot\left(\frac{x}{2}\right)>0$ we have $$\arg\left(i\cot\left(\frac{x}{2}\right)\right)=\frac{\pi}{2}$$ then finally $$\sum_{n\geq1}\frac{n}{4n^{2}-1}\sin\left(2nx\right)=\color{red}{\frac{\pi\cos\left(x\right)}{8}},\,2\pi k<x<2\pi x+\pi,\,k\in\mathbb{Z}.$$

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  • $\begingroup$ Thanks, this isn't part of the question, though maybe it should be: evaluating the right hand side of the series at $x=0$ is identically $0$, yet $\cos(0) = 1$... How is this expansion possible? $\endgroup$ – jaslibra Jul 17 '17 at 17:12
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    $\begingroup$ @jaslibra Take a look at the domain of $x$, you may see that $x=0$ is not acceptable. $\endgroup$ – Marco Cantarini Jul 17 '17 at 19:12

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