Geometric series of $\frac{1}{1+3e^{i\,t}}$

Question

What's the geometric sum equivalent to the rational expression shown below? $$\frac{1}{ 1+3e^{i\,t}},\quad t\in\mathbb{R^+}$$

I've been trying for so long, and Wolfram|Mathematica won't help me either.

Answer 1

Write $$\frac{1}{1+3e^{it}}=\frac{1}{3e^{it}}\frac{1}{1+\frac{1}{3e^{it}}}$$

Then, we have

$$\frac{1}{1+3e^{it}}=\frac{1}{3e^{it}}\sum_{n=0}^\infty\left(-3e^{it}\right)^{-n}=\sum_{n=1}^\infty (-1)^{n+1}\left(\frac{1}{3}\right)^ne^{-int}$$

Answer 2

For $|x| \lt 1$ it holds that $\sum_{k=0}^\infty x^k = \frac{1}{1-x}$, and therefore $\sum_{k=0}^\infty (-3x)^k = \frac{1}{1+3x}$ (if $|x| \lt \frac13$).

Answer 3

Hint. You have, for $|z|<\dfrac13$,

$$ \frac{1}{ 1+3z}=\sum_{n=0}^{\infty}(-1)^n3^n z^n $$

Answer 4

$$ \frac{1}{1+3x}=\sum_{n=0}^\infty(-3)^nx^n \quad |x|<\frac13. $$