Find the Sum of the Series Using Complex Exponentials

Find the sum of the series $$\sum_{n=0}^{\infty}\frac{\cos(nx)}{2^{n}}$$ and $$\sum_{n=0}^{\infty}\frac{\sin(nx)}{2^{n}}$$.

Hint: Rewrite the trigonometric functions using complex exponentials.  This is what I have so far:

$$\cos(nx) = \frac{e^{inx}+e^{-inx}}{2}$$ and $$\sin(nx) = \frac{e^{inx}-e^{-inx}}{2i}$$

By substituting in the values into the given series, we get:

$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{e^{inx}+e^{-inx}}{2}$$ + $$\frac{1}{2}\sum_{n=0}^{\infty}\frac{e^{inx}-e^{-inx}}{2i}$$.

Edit: used latex codes \cos and \sin.

• Use the $C+iS$ method of summation, it will be easier. – Awe Kumar Jha Nov 7 '18 at 3:33
• @AweKumarJha I'm confused by what you're suggesting – AmR Nov 7 '18 at 3:41

First, $$e^{inx} = \cos(nx)+i\sin(nx)$$.
So, $$\displaystyle\sum_{n=0}^\infty \left(\frac{\cos(nx)}{2^n}+i\frac{\sin(nx)}{2^n}\right) = \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{2^n} = \displaystyle\sum_{n=0}^\infty \left(\frac{e^{ix}}{2}\right)^n=\frac{2}{2-e^{ix}}$$.
• @AmR somewhat wrong because the denominators should be $2^n$. Say you have that denominator, then it is not wrong, although you need to compute for both $\sum \frac{e^{inx}}{2^n}$ and $\sum \frac{e^{-inx}}{2^n}$, and do it twice; thus, more work. – Isko10986 Nov 7 '18 at 19:24