For the input $x[n]=1$ for all $n$, how to get to this without any shortcuts? (I know we can use the "well-known" property of the periodic signals) I want to reach this final form using pure mathematics.
$$ X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} = \sum_{n=-\infty}^{+\infty} e^{-j\omega n}=\sum_{k=-\infty}^{+\infty} 2\pi \delta(\omega +2\pi k)$$
If fact, I want to know why we get:
$$ \sum_{n=-\infty}^{+\infty} e^{-j\omega n}=\sum_{k=-\infty}^{+\infty} 2\pi \delta(\omega +2\pi k)$$
Why the summation of the complex envelopes (the exponential one) can be expressed as the summation of the impulse function (the Dirac delta one)? How are they related?
Can I answer this question without resorting to the Fourier analysis?