# Infinite sum

#### jacobi

##### Active member
Find $$\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$.

#### M R

##### Active member
Find $$\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$.

By considering the real part of $$\displaystyle \left(\frac{e^{ix}}{2}\right)^n$$ and summing a GP I get $$\displaystyle \frac{4-2\cos(x)}{5-4\cos(x)}$$...I think.

#### chisigma

##### Well-known member
Find $$\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$.
Applying the Euler's identity...

$\displaystyle \cos (n x) = \frac{e^{i n x} + e^{- i n x}}{2}\ (1)$

... You obtain...

$\displaystyle \sum_{n =0}^{\infty} \frac{\cos (n x)}{2^{n}} = \frac{1}{2}\ \sum_{n =0}^{\infty} (\frac{e^{i x}}{2})^{n} + \frac{1}{2}\ \sum_{n =0}^{\infty} (\frac{e^{- i x}}{2})^{n} =\frac{1}{2}\ \frac{1}{1 - \frac{e^{i x}}{2}} + \frac{1}{2}\ \frac{1} {1 - \frac{e^{- i x}}{2}}\ (2)$

The task of 'suppression' of the imaginary terms from the (2) is tedious but not very difficult and is left to You...

Kind regards

$\chi$ $\sigma$