How can we prove that $e^{-jωn}$ converges at $0$ while n -> infinity?

Question

I am trying to find the DTFT of a function.

Here is my problem:

How does he prove that $e^{(-jω)m+1} = 0$ ?

Answer 1

This is just a formal answer to the question (it has already been answered in pieces in the above comments).

Let $z=a+bj$ be a complex number (where $a, b$ are real numbers and $j = \sqrt{-1}$). The magnitude of $z$ is the distance between $(a,b)$ and $(0,0)$ in the 2-d plane: $$ |z| = \sqrt{a^2 + b^2}$$

In general we have the geometric sum formula for a finite sum: $$ \sum_{i=0}^{n}z^i = \frac{1-z^{n+1}}{1-z} $$ If $|z|<1$ then $|z|^n\rightarrow 0$. So, taking a limit as $n\rightarrow\infty$ gives the infinite sum formula in the special case when $|z|<1$: $$ \boxed{\sum_{i=0}^{\infty} z^i = \frac{1}{1-z} \quad , \quad \mbox{(if $|z|<1$)}} $$ In your summation you have (assuming $\omega \in \mathbb{R}$): $$ z = \frac{e^{-j\omega}}{4} = \frac{cos(\omega) - j\sin(\omega)}{4} $$ $$ \implies |z| = \frac{1}{4} < 1 $$ so the infinite sum formula (in the above box) can be applied.