# Infinite Sum

1. Mar 19, 2006

### cepheid

Staff Emeritus
I'm trying to figure out how to do this sum and not sure where to start:

$$\sum_{n = -\infty}^{-1} \left(\frac{1}{2}e^{-j \omega} \right)^n$$

I tried rewriting it like this:

$$\sum_{m = 1}^{\infty} (2e^{j \omega})^m$$

m = -n

but I'm not sure if that helps.

2. Mar 19, 2006

### 0rthodontist

What are j and omega? Unless I'm missing something you have an infinite power series with ratio r = 2e^(j * omega) whose sum can be found by the formula r/(1-r).

3. Mar 19, 2006

### d_leet

That will work provided that (j*omega) < ln(1/2) because in order for an infinite geometric series to converge the common ratio must be less than 1.

4. Mar 19, 2006

### 0rthodontist

True.
whitespace

5. Mar 19, 2006

### cepheid

Staff Emeritus
Tsk tsk. You guys don't instantly recognize the combination jω?

$$j = \sqrt{-1}$$

I know I know. Those crazy electrical engineers...and omega is of course the angular frequency. What you are looking at is the calculation of the Fourier transform of a discrete time signal x[n], n in the set of integers.

Anyway, you must be wondering, if that's a complex exponential, then how could the series possibly converge? Well it's a moot point, because I made a mistake. Originally I had:

$$X(e^{j \omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j \omega n} = \sum_{n=-\infty}^{-1}\left(\frac{1}{2}\right)^{-n}e^{-j \omega n}$$

Which I should have expressed like so:

$$\sum_{n=-\infty}^{-1}\left[\left(\frac{1}{2}\right)^{-1}e^{-j \omega }\right]^n$$

which is decidedly different from what I started out with in my first post. It becomes (setting m = -n):

$$\sum_{m=1}^{\infty}\left(\frac{1}{2}e^{j \omega }\right)^m = \frac{\frac{1}{2}e^{j \omega }}{1 - \frac{1}{2}e^{j \omega }}$$

Correct?