You don't actually get 2 \pi at \omega = 0, you get infinity. You get 2 \pi \delta ( \omega ) evaluated at \omega = 0, which is infinite or not well-defined at least.
Yes, that is correct. You should get 0 when \omega \neq 2 \pi k for any integer k.
Do you know how to calculate the...
\displaystyle \sum_{n=-\infty}^{\infty} e^{-jn \omega} = \sum_{n=0}^{\infty} e^{-jn \omega} + \sum_{n=-\infty}^{0} e^{-jn \omega} -1 = \sum_{n=0}^{\infty} e^{-jn \omega} + \sum_{n=0}^{\infty} e^{jn \omega} - 1.
When the sum converges, each of the last two sums is a geometric series that you...
There is something slightly messy about the proof of orthogonality given by maverick6664. Technically, one should show that the boundary terms that result from each integration by parts vanish. This is fairly straightforward to do, but there's a much cleaner way of proving orthogonality which...