If $e^{i\theta }=\mathrm{cis}\theta$ then $\sum _{n=0}^{\infty }\frac{\cos \left(n\theta \right)}{2^n}=$

Let

$C=\sum _{n=0}^{\infty }\frac{\cos n\theta }{2^n}$

$\Rightarrow C=1+\frac{\cos \left(\theta \right)}{2}+\frac{\cos \left(2\theta \right)}{2^2}+\frac{\cos \left(3\theta \right)}{2^3}+....$

$iS=\frac{i\sin \left(\theta \right)}{2}+\frac{i\sin \left(2\theta \right)}{2^2}+\frac{i\sin \left(3\theta \right)}{2^3}+....$

Then,

$C+iS=1+\frac{1}{2}\left(\cos \theta +i\sin \theta \right)+\frac{1}{2^2}\left(\cos 2\theta +i\sin 2\theta \right)+\frac{1}{2^3}\left(\cos 3\theta +i\sin 3\theta \right)+....$

$\Rightarrow C+iS=1+\frac{1}{2}{e}^{i\theta }+\frac{1}{2^2}{e}^{2i\theta }+\frac{1}{2^3}{e}^{3i\theta }+......$

$\Rightarrow C+iS=1+\frac{\frac{1}{2}{e}^{i\theta }}{1-\frac{1}{2}{e}^{i\theta }}$

$\Rightarrow C+iS=1+\frac{e^{i\theta }}{2-e^{i\theta }}$

$\Rightarrow C+iS=1+\frac{\cos \theta +i\sin \theta }{2-\cos \theta -i\sin \theta }$

$\Rightarrow C+iS=1+\left[\left(\frac{\cos \theta +i\sin \theta }{2-\cos \theta -i\sin \theta }\right)\times \left(\frac{2-\cos \theta +i\sin \theta }{2-\cos \theta +i\sin \theta }\right)\right]$

$\Rightarrow C+iS=1+\left[\frac{\cos \theta \left(2-\cos \theta +i\sin \theta \right)+i\sin \theta \left(2-\cos \theta +i\sin \theta \right)}{{\left(2-\cos \theta \right)}^2+{\sin }^2\theta }\right]$

$\Rightarrow C+iS=1+\left[\frac{2\cos \theta -{\cos }^2\theta -{\sin }^2\theta +2i\sin \theta }{5-4\cos \theta }\right]$

$\Rightarrow C+iS=1+\left[\frac{2\cos \theta -1+2i\sin \theta }{5-4\cos \theta }\right]$

$\Rightarrow C+iS=\left[\frac{4-2\cos \theta +2i\sin \theta }{5-4\cos \theta }\right]$

$\Rightarrow C+iS=\left(\frac{4-2\cos \theta }{5-4\cos \theta }\right)+i\left(\frac{2\sin \theta }{5-4\cos \theta }\right)$

On comparing real and imaginary part, we get

$C=\frac{4-2\cos \theta }{5-4\cos \theta }$