This page will show how to compute the Fourier transforms of CT and DT signals that have a power of absolute value (e.g. $ (\frac{1}{2})^{|n|} $). First, I will show an example of Professor Mimi's, then I will solve a different problem.
$ x[n] = (\frac{1}{2j})^(|n|) $
$ X(w) = \sum_{n=-\infty}^{\infty} x[n]e^{-jwn} $
$ = \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{|n|}e^{-jwn} $
You can think of |n| in two different ways, but they both reach the same conclusion.
1) Sum x[n]u[n], and x[n]u[-n-1].
$ X(w) = \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{-n}u[-n-1]e^{-jwn} + \sum_{n=-\infty}^{\infty} (\frac{1}{2j})^{n}u[n]e^{-jwn} $
$ X(w) = \sum_{n=-\infty}^{-1} (\frac{1}{2j})^{-n}e^{-jwn} + \sum_{n=0}^{\infty} (\frac{1}{2j})^{n}e^{-jwn} $
Since there a u[n] functions in this method, it might be a little easier to set the bounds of the summation.
2) Or, just come to the conclusion that |n| = n if n>=0, and |n| = -n if n<0.
Now, since we have bounds for the summation, we can continue to solve the two summations and come to the solution that...
$ X(w) = \frac{1}{1-\frac{e^{jw}}{2j}} - 1 + \frac{1}{1-\frac{1}{2je^{jw}}} $
