This is a rather simple question, but I'm stuck on one step. Here's what I've done:
1) $x[n] = (n+2)(\frac{1}{2})^nu[n] = n(\frac{1}{2})^nu[n]+2(\frac{1}{2})^nu[n]$
The discrete fourier transform is defined as $X(e^{-i\omega}) = \sum\limits_{n=-\infty}^{\infty} x[n]e^{-i\omega n}$
2) Therefore, $X(e^{-i\omega}) = \sum\limits_{n=-\infty}^{\infty} (n(\frac{1}{2})^nu[n]+2(\frac{1}{2})^nu[n])e^{-i\omega n} = \sum\limits_{n= 0}^{\infty} (n(\frac{1}{2})^n+2(\frac{1}{2})^n)e^{-i\omega n}$
The second term in the sum is a geometric series that converges to $\frac{2}{1-\frac{1}{2}e^{-i\omega}}$.
However, I'm stuck on the first term. What does $\sum\limits_{n= 0}^{\infty} n(\frac{1}{2})^ne^{-i\omega n}$ converge to?
I have a solution that says it converges to $\frac{\frac{1}{2}e^{-i\omega}}{(1-\frac{1}{2}e^{-i\omega})^2}$.
If the solution is correct, how does one arrive at it? In other words, what type of series is it and what is the proof of its convergence?
If you have $\sum_{n=0}^{\infty} n ~ a^n$ you can treat this with some trickery as: $$a \sum_{n=0}^{\infty} n ~ a^n = a ~ \frac{d}{da} \sum_{n=0}^{\infty} a^n = a ~ \frac{d}{da} \frac {1}{1-a} = \frac a {(1 - a)^2}.$$ Repeat as necessary for any polynomial you wish.