Interval of convergence of trig. series involving complex numbers

Question

I have been working on the following problem:

Determine the sum of the convergent trigonometric series:

$$ \sum_{k=-\infty}^{\infty} \frac{i}{3^{|k|}}e^{ikx} $$

This is my work so far: Because of the absolute value in the above expression, we can rewrite it.

$$ \sum_{k=0}^{\infty} \frac{i}{3^{k}} \left(e^{ikx} + e^{-ikx} \right) = \sum_{k=0}^{\infty} \frac{i}{3^{k}} e^{ikx} + \sum_{k=1}^{\infty} \frac{i} {3^{k}} e^{-ikx} $$

Now let's look at the first summand. We can rewrite this to get a geometric series:

$$ \frac{i}{3^{k}} e^{ikx} = i \frac{e^{ikx}}{3^{k}} = i \left( \frac{e^{ix}}{3} \right)^k $$

And thereby:

$$ \sum_{k=0}^{\infty} i \left( \frac{e^{ix}}{3} \right)^k = \frac{i}{\frac{3}{3}-\frac{e^{ix}}{3}} = \frac{3i}{3-e^{ix}} $$

when $ \left| \frac{e^{ix}}{3} \right| < 1 $. And likewise with the second summand which is convergent for $ \left| \frac{e^{-ix}}{3} \right| < 1 $

What worries me is the expression $e^{ix}$ and $e^{-ix}$ in the above inequality. I would like to find the interval of convergence, or I guess, the disk of convergence for the series, and maybe plot it. I guess in this case it would be the intersection of two disks.

How would one do this? How would one know where the center is?

Answer 1

Hint

$$ \sum_{k=-\infty}^{\infty} \frac{i}{3^{|k|}}e^{ikx} = i\left(1+\sum_1^{\infty}\left(\left(\frac{e^{i x}}{3}\right)^k+\left(\frac{e^{-i x}}{3}\right)^k\right)\right) = i\left(\frac{3}{3+e^{ix}}+\frac{3}{3+e^{-ix}}-1\right) $$

We can use also de Moivre's identity.

Answer 2

Okay - so here's a new attempt using some of the hints. Rewriting once again we have:

$$ \sum_{k=-\infty}^{\infty} \frac{i}{3^{|k|}} e^{ikx} = i \left( 1 + \sum_{k=1}^{\infty} \left( \frac{e^{ix}}{3} \right)^k + \sum_{k=1}^{\infty} \left( \frac{e^{-ix}}{3} \right)^k \right) $$

So here's what I don't understand. Giving we have to convergent geometric series I would say that the sum is:

$$ i \left( 1 + \sum_{k=1}^{\infty} \left( \frac{e^{ix}}{3} \right)^k + \sum_{k=1}^{\infty} \left( \frac{e^{-ix}}{3} \right)^k \right) = i \left( 1 + \frac{\frac{e^{ix}}{3}}{1 - \frac{e^{ix}}{3}} + \frac{\frac{e^{-ix}}{3}}{1 - \frac{e^{-ix}}{3}} \right) $$

where the formula for the sum of a geometric series have been used. That is $\frac{r}{1-r}$, since we are starting at $k=1$. So the sum is:

$$ i \left( 1 + \frac{e^{ix}}{3 - e^{ix}} + \frac{e^{-ix}}{3 - e^{-ix}} \right) $$

But this doesn't agree with Cesareo's answer. What am I missing here?