Show that $\displaystyle \sum_{ n \in \mathbb{Z} \setminus \{0\}}^{} \frac{e^{2 \pi i xn}}{n}$ is a convergent series for all real x. I think that this could be done by breaking up the sum into two pieces; the piece that has n going to $\infty$ and the other going to $-\infty$. From there I have applied that fact that $e^{2 \pi x i n}= \cos(2\pi x n)+ i\sin(2 \pi x n)$ and then the fact that $\displaystyle \frac{\cos(2\pi x n)}{n}+ i\frac{\sin(2 \pi x n)}{n}$ is a sequence such that the series of $\displaystyle \frac{\cos(2\pi x n)}{n}$ and $\displaystyle \frac{\sin(2 \pi x n)}{n}$ both have convergent series. The thing is that I only know these two series are convergent when n goes from 1 to infinity. Is it still true in going from -1 to - infinity? Does the method of this proof even work? Thanks!
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1I edited some of the TeX. Please check if everything is correct. – Git Gud Jan 24 '13 at 23:08
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3$\cos(2\pi n)$ is just $1$. $\sin(2\pi n)$ is just $0$. What am I missing? – mjqxxxx Jan 24 '13 at 23:09
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yea for the negative values, you can take minus before the sum – Guest 86 Jan 24 '13 at 23:09
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3Plus, unless someone tells you the order in which to sum the terms, this sum isn't well-defined (because it doesn't converge absolutely). – mjqxxxx Jan 24 '13 at 23:11
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Ah!sorry I forget the "x" I'm gonna adjust the question. – Jmaff Jan 25 '13 at 00:07
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I should mention that I only know that $\sum_{n=1}^{\infty} cos(2\pi x n )/n $ and $\sum_{n=1}^{\infty} sin(2\pi x n )/n $ converge because of wolfram alpha. I am trying to prove them myself though. – Jmaff Jan 25 '13 at 00:49
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possible duplicate of Give a demonstration that $\sum\limits_{n=1}^\infty\frac{\sin(n)}{n}$ converges. – Feb 02 '13 at 20:17
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Since $$\frac{e^{i2\pi nx}}{n}+\frac{e^{-i2\pi nx}}{-n}=2i\frac{\sin(2\pi nx)}{n}$$ your series is equal to $$2i\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{n}=4\pi i\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{2n\pi}.$$ But $\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{2n\pi}$ is the Fourier sine series of $$f(x)=\frac{1}{2}\left(\frac{1}{2}-x\right), 0\le x\le 1/2.$$ Let $F(x)$ be the odd period 1 extension of f(x). By convergence of Fourier series, the series $\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{2n\pi}$ converges to $\frac{F(x-)+F(x+)}{2}$ for all $x$.
You can also use Dirichlet Test for convergence.
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