So here is my problem,
I would like to prove an identity I found in a book which was given without a proof.
Namely $$-i\sum_{n\in\mathbb Z} \operatorname{sign}(n)\cdot e^{i2\pi nx}=\cot(\pi x)$$ I thought since there is no explicit proof it would be easy to show it but I failed...
Could someone help me? Or does anybody know where I can find the proof?
Thank you!
Partial Sums Oscillate
One problem making this sum difficult is that it doesn't converge because the terms do not tend to $0$. Using the formula for the sum of a geometric series, we get $$ \begin{align} -i\left(\sum_{k=0}^ne^{2\pi ikx}-\sum_{k=0}^ne^{-2\pi ikx}\right) &=-i\left(\frac{1-e^{2\pi i(n+1)x}}{1-e^{2\pi ix}}-\frac{1-e^{-2\pi i(n+1)x}}{1-e^{-2\pi ix}}\right)\\ &=-i\left(\frac{e^{-\pi ix}-e^{\pi i(2n+1)x}}{e^{-\pi ix}-e^{\pi ix}}-\frac{e^{\pi ix}-e^{-\pi i(2n+1)x}}{e^{\pi ix}-e^{-\pi ix}}\right)\\[3pt] &=\frac{\cos(\pi x)-\cos((2n+1)\pi x)}{\sin(\pi x)}\\[3pt] &=\cot(\pi x)-\frac{\cos((2n+1)\pi x)}{\sin(\pi x)}\tag1 \end{align} $$ So the partial sums oscillate around $\cot(\pi x)$, but they never converge (as expected).
Another Method of Summation
However, partial sums are not the only way to sum Fourier series. Partial sums correspond to the Dirichlet Kernel applied to the function. The Dirichlet Kernel is an approximation to the delta function, but it is not in $L^1$. As such, it does not approximate as well as the Fejér Kernel, which is in $L^1$ (in fact, the Fejér Kernel is non-negative). As the Dirichlet Kernel corresponds to partial summation, the Fejér Kernel corresponds to Cesàro Summation, which is like an average of partial sums.
Applying the Dirichlet Kernel corresponds to multiplying the Fourier coefficients by $[\,|k|\le n\,]$ ($\,[\dots]$ are Iverson brackets). Dirichlet Kernel approximations are subject to Gibb's Phenomenon around jump discontinuities (see this answer for details on the ringing).
Applying the Fejér Kernel corresponds to multiplying the Fourier coefficients by $\left(1-\frac{|k|}n\right)[\,|k|\le n\,]$.
Lemma: $$ \sum_{k=0}^n\left(1-\frac kn\right)x^k=\frac{n-(n+1)x+x^{n+1}}{n(1-x)^2}\tag2 $$ Proof:
Since the terms with $x^k$ for $2\le k\le n$ cancel, we get $$ \begin{align} (1-x)^2\sum_{k=0}^n(n-k)x^k &=\sum_{k=0}^n(n-k)x^k-2\sum_{k=0}^n(n-k)x^{k+1}+\sum_{k=0}^n(n-k)x^{k+2}\\ &=\underbrace{\sum_{k=0}^n(n-k)x^k}_{k\in\{0,1\}}-2\underbrace{\sum_{k=1}^{n+1}(n-k+1)x^k}_{k\in\{1,n+1\}}+\underbrace{\sum_{k=2}^{n+2}(n-k+2)x^k}_{k\in\{n+1,n+2\}}\\ &=n-(n+1)x+x^{n+1}\tag3 \end{align} $$ $\square$
Therefore, the Fejér Kernel applied to the sum in question can be evaluated as $$ \begin{align} &-i\left(\sum_{k=0}^n\left(1-\frac kn\right)e^{2\pi ikx}-\sum_{k=0}^n\left(1-\frac kn\right)e^{-2\pi ikx}\right)\\ &=-i\left(\frac{n-(n+1)e^{2\pi ix}+e^{2\pi i(n+1)x}}{n\left(1-e^{2\pi ix}\right)^2} -\frac{n-(n+1)e^{-2\pi ix}+e^{-2\pi i(n+1)x}}{n\left(1-e^{-2\pi ix}\right)^2}\right)\\ &=-i\left(\frac{ne^{-2\pi ix}-(n+1)+e^{2\pi inx}}{n\left(e^{-\pi ix}-e^{\pi ix}\right)^2} -\frac{ne^{2\pi ix}-(n+1)+e^{-2\pi inx}}{n\left(e^{\pi ix}-e^{-\pi ix}\right)^2}\right)\\[3pt] &=-i\frac{2ni\sin(2\pi x)-2i\sin(2\pi nx)}{4n\sin(\pi x)^2}\\[6pt] &=\cot(\pi x)-\frac{\sin(2\pi nx)}{2n\sin(\pi x)^2}\tag4 \end{align} $$ which does converge to $\cot(\pi x)$.
