Series with rotating phase

[hilbert2]

TL;DR

About series where the complex phase of the terms rotates around the unit circle.

It's a basic fact that the harmonic series

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$

diverges, while the similar series with alternating sign

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$

converges to the value $\log 2$.

If I compute partial sums for the series

$\displaystyle\sum_{n=1}^{\infty}\frac{e^{ic(n-1)}}{n}$

where $c$ is a real-valued constant, the real and imaginary parts seem to approach a finite value for any nonzero value of $c$. The case $c=\pi$ is just the series with alternating sign.

Do these series with rotating phase factor appear in any applications? It seems to be a generalization of the concept whether a series of real-valued terms is alternating or has only terms of same sign.

 

[anuttarasammyak]

It is a Fourier series which has abundant applications. it has relation with delta function.

[EDIT]
Though a bit different from your formula, complex Fourier series for Heaviside step function, H(x)=0 for x<0 H(0)=1/2 H(x)=1 for x>0 in period 2$\pi$, is
$$H(x)=\frac{1}{2}+\frac{1}{\pi}\sum_{k=-\infty\ odd}^{+\infty}\frac{e^{ikx}}{ik}$$
With period $2\pi$ this is a square wave.

$$H'(x)= \frac{1}{\pi}\sum_{k=-\infty\ odd}^{k=\infty}e^{ikx}= \sum_{n=-\infty}^{\infty} [\delta(x-2n\pi )-\delta(x-(2n+1)\pi)]$$
where
$$\delta(x)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty} e^{ikx}$$

 

[hilbert2]

A good point, I didn't think of it as defining a function of $c$, just as a series that produces a number when summed.

The sum of that rotating phase harmonic series seems to be

$\displaystyle\sum_{n=0}^{\infty}\frac{e^{i(n-1)x}}{n} = e^{-ix}\log (1-e^{ix})$,

which clearly diverges when $x\rightarrow 0$ and has a finite value for $x>0$ no matter how close to zero $x$ is.

Here's a discussion about something that can be interpreted as similar but with accelerating phase rotation: https://math.stackexchange.com/questions/2229263/convergence-of-sum-frac-sinn2n

 

[julian]

A good point, I didn't think of it as defining a function of $c$, just as a series that produces a number when summed.

The sum of that rotating phase harmonic series seems to be

$\displaystyle\sum_{n=0}^{\infty}\frac{e^{i(n-1)x}}{n} = e^{-ix}\log (1-e^{ix})$,

which clearly diverges when $x\rightarrow 0$ and has a finite value for $x>0$ no matter how close to zero $x$ is.

Here's a discussion about something that can be interpreted as similar but with accelerating phase rotation: https://math.stackexchange.com/questions/2229263/convergence-of-sum-frac-sinn2n

You can prove $\ln (1-z)$ converges everywhere on the unit circle except at $z=1$ by Dirichlet's test: Fix $z$ in the unit circle, i.e. $|z|=1$.

If $\{ a_n \}$ are real numbers and $\{ b_n \}$ complex numbers such that: (i) $a_1 \geq a_2 \geq \cdots$ (ii) $\lim_{n \rightarrow \infty} a_n = 0$ (iii) There exists $M$ such that $\left| \sum_{n=1}^N b_n \right| \leq M$ for all $N \in \mathbb{N}$; then $\sum_{n=1}^\infty a_n b_n$ converges.

Choose $a_n = 1/n$ and $b_n = z^n$, then the first two conditions obviously hold. For the third condition:

\begin{align*}
\left| \sum_{n=1}^N z^n \right| = \left| \dfrac{z-z^{N+1}}{1-z} \right| \leq \frac{2}{| 1-z |}
\end{align*}

for all $N \in \mathbb{N}$. This shows the series converges for $|z|=1$ where $z \not= 1$.