# Show that $\sum\limits_{ n \in \mathbb{Z} \setminus \{0\}}\frac{e^{2 \pi i x n}}{n}$ is a convergent series where $x$ is any real

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!

• I edited some of the TeX. Please check if everything is correct. – Git Gud Jan 24 '13 at 23:08
• $\cos(2\pi n)$ is just $1$. $\sin(2\pi n)$ is just $0$. What am I missing? – mjqxxxx Jan 24 '13 at 23:09
• yea for the negative values, you can take minus before the sum – Guest 86 Jan 24 '13 at 23:09
• Plus, 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
• Ah!sorry I forget the "x" I'm gonna adjust the question. – Jmaff Jan 25 '13 at 0:07

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 $x$.
Hint Consider $\dfrac{e^{2\pi ixn}}{n}+\dfrac{e^{-2\pi ixn}}{-n}$