Determining convergence of $P(z)=1-\exp(i\theta)+\exp^{2}(i\theta)+\dots$

Question

Given the geometric series $$P(z)=1-\exp(i\theta)+\exp^{2}(i\theta)-\exp^{3}(i\theta)+\dots$$ where $z$ represents a complex number, I would like to determine the values of $\theta$ such that $P(z)$ converges.

From the geometric ratio $r=-\exp(i\theta)$, I determined that $\lvert r \rvert=1$. so I used the alegbraic formula for sum to infinity $S_N$ of a geometric series to obtain $$S_N=\frac{1-(-\exp(i\theta))^N}{1-(-\exp(i\theta))}, $$ where $N\to \infty$.

Using this expression, how should I proceed to find the values of $\theta$ such that $S_N$ converges?

Edit: Most of the people that replied says that the series is divergent. However, according to the textbook 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence, the above series converges except at $\theta =\pi$. I've posted the screenshots here and here.

Answer 1

First method

So you your have geometric sum:

$$ S_N= \frac{1+(-1)^Ne^{i\theta N}}{1+e^{i\theta}} $$

So you only need

$$ \kappa_N\triangleq \Re((-1)^Ne^{i\theta N})= \cos((\theta+\pi)N) $$

convergent

Suppose it converges to $l$:

$$ (\theta+\pi)N = \arccos(l ) + o(1)$$

So necesserely:

$$ \theta = -\pi \mod[2\pi] $$

Idem for $ \Im $ Part

But your quotient isn't define so it doesn't works. Never convergent.

Second Method

Your general term doesn't tend to zero so divergent series it involves !

Answer 2

Since $\theta\in\mathbb{R}$, $\bigl|\exp(i\theta)\bigr|=e^{\operatorname{Re}(i\theta)}=1$ and therefore your series never converges, since you don't have $\lim_{n\to\infty}(-1)^n\exp(i\theta)^n=0$.