Decomposition of a series

Question

I was studying mathematical series and I stumbled across a really useful example but there is a part I'm not really sure about.

Basically, we have been given two series that are the real and imaginary part of a sequence $w_n = (\frac{e^i}{2})^n$ so we can say that the two previous series converge since $w_n$ converge.

My issue is that in the correction they assume $\sum_{n=0}^{\inf+}w_n = \frac{1}{1-\frac{e^i}{2}}$

I'm a little confused about this. I guess since it's a sum we can use the formula: $\frac{1-q^{n+1}}{1-q}$ but why does it simplify like $\frac{1}{1-\frac{e^i}{2}}$ Are we just saying that since n tends to infinity, $(\frac{e^i}{2})^{n+1}$ tends to $0$ and just like that get rid of it or is there a more mathematical way?

Answer 1

By definition, $\sum_{n=0}^\infty\left(\frac{e^i}2\right)^n$ means $\lim_{N\to\infty}\sum_{n=0}^N\left(\frac{e^i}2\right)^n$. But\begin{align}\lim_{N\to\infty}\sum_{n=0}^N\left(\frac{e^i}2\right)^n&=\lim_{N\to\infty}\frac{1-\left(\frac{e^i}2\right)^{N+1}}{1-\frac{e^i}2}\\&=\frac1{1-\frac{e^i}2},\end{align}since $\lim_{N\to\infty}\left(\frac{e^i}2\right)^{N+1}=0$, which follows from the fact that $\left|\frac{e^i}2\right|=\frac12$.