Evaluate $\sum\limits_{n=0}^{\infty} \frac{\cos(nx)}{2^n}$ where $\cos x = \frac15$ Evaluate 
$$\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{2^n}$$
where $\cos x = \frac{1}{5}$.
This is a complex number question. But I don’t know where to start. Maybe need to use the DeMoivre’s Theorem?
 A: Consider the following auxiliar sum: $\sum_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}$, and use that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ to get: $\sum_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+\sum_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+i\dfrac{\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{\cos(nx)+i\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{e^{inx}}{2^{n}}=\sum_{n=0}^\infty (\dfrac{e^{ix}}{2})^{n}=\dfrac{1}{1-e^{ix}/2}=\dfrac{2}{2-e^{ix}}=\dfrac{2}{2-\cos(x)-i\sin(x)}=\dfrac{2}{((2-\cos(x))-i\sin(x))}\dfrac{(2-\cos(x))+i\sin(x)}{((2-\cos(x))+i\sin(x))}=\dfrac{4-2\cos(x)+2i\sin(x)}{(2-\cos(x))^{2}+\sin^{2}(x)}=\dfrac{4-2\cos(x)+2i\sin(x)}{4-4\cos(x)+\cos^{2}(x)+\sin^{2}(x)}=\dfrac{4-2\cos(x)+2i\sin(x)}{5-4\cos(x)}=\dfrac{4-2\cos(x)}{5-4\cos(x)}+i\dfrac{2\sin(x)}{5-4\cos(x)}$
and comparing imaginary and real parts with $\sum\limits_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+\sum\limits_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}=\sum\limits_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+i\sum\limits_{n=0}^\infty \dfrac{\sin(nx)}{2^{n}}$ 
to obtain: $\sum\limits_{n=0}^\infty \dfrac{\sin(nx)}{2^{n}}=\dfrac{2\sin(x)}{5-4\cos(x)}$ and $\sum\limits_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}=\dfrac{4-2\cos(x)}{5-4\cos(x)}$
A: $\frac {\cos (nx)} {2^{n}}$ is the real part of $(\frac {e^{ix}} 2)^{n}$. Thus you only have to sum a geometric series.
A: If $S=\sum_{r=0}^\infty y^r\cos r x$
Using scale of relation from this or $\#187$ of this(downloadable)
$$(1-2y\cos x+y^2)S$$
$$=1+y\cos x+y^2\cos2x+y^3\cos3x+\cdots$$
$$-2y\cos x-2y^2\cos^2x-2y^3\cos2x\cos x-2y^4\cos3x\cos x-\cdots$$
$$+y^2+y^3\cos x+y^4\cos2x+y^5\cos3x+\cdots$$
$$=1+y(-\cos x)+y^2(\cos2x-2\cos^2x+1)+y^3(\cos3x-2\cos3x\cos x+\cos x)+\cdots$$
$$\implies S=\dfrac{1-y\cos x}{1-2y\cos x+y^2}$$
Here $2y=1$
