# Decomposition of a series

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?

• Are you sure that it's $\left(\frac{e^i}{2^n}\right)^n$ and not just $\left(\frac{e^i}2\right)^n$? Oct 26, 2021 at 13:07
• Oh yeah sorry and thanks for noticing Oct 26, 2021 at 13:08
• As $n\to+\infty$, $\left|\left(\frac{e^{i}}{2}\right)^{n+1}\right|=\frac{1}{2^{n+1}}\to 0$. I believe, this is mathematical way. Oct 26, 2021 at 13:14

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 $$\left|\frac{e^i}2\right|=\frac12$$.