I would like to study the series
$$ \sum_{n\in\mathbb{N}}\left(\frac{e^{it}}{2}\right)^{n} $$
My idea is to first look at what can be said concerning any type of convergence:
$$ \forall{t}\in\mathbb{R},\forall n\in\mathbb{N} : \left\lvert\left(\frac{e^{it}}{2}\right)^{n}\right\rvert = \frac{1}{2^{n}}\left\lvert e^{itn}\right\rvert = \frac{1}{2^{n}}(\cos^{2}(tn) + \sin^{2}(tn))^{1/2}=\frac{1}{2^{n}}\tag{1}\label{1} $$
Thus we conclude that the series is absolutely convergent :
$$ \forall t\in\mathbb{R} : \sum_{n\in\mathbb{N}}\left\lvert\left(\frac{e^{it}}{2}\right)^{n}\right\rvert = \sum_{n\in\mathbb{N}}\frac{1}{2^{n}} = 2 $$
Which implies the pointwise convergence of the series and the normal convergence can also be deduced from $\eqref{1}$.
However, I did not succeed to find the limit, anyone has an idea ?
Thank you a lot
Hint: I presume you know that $$\sum_{n = 0}^\infty z^n = \frac{1}{1 - z}$$ for $\left|z\right| < 1$. Now consider $z = \frac{e^{it}}{2}$.