# Proving DTFT pair as special case of another.

Consider these 2 basic discrete-time Fourier transform (DTFT) pairs...

$$\require{extpfeil}\Newextarrow{\xleftrightarrow}{15,15}{0x2194} \begin{array}{rcl} u[n] & \xleftrightarrow{\mathscr{F}} & \frac{1}{1 - \text{e}^{-j\omega}} + \sum_{k=-\infty}^{\infty}\left[\pi\delta\left[\omega - 2\pi k\right]\right] \\ \frac{(n+r-1)!}{n!(r-1)!}a^nu[n] & \xleftrightarrow{\mathscr{F}} & \frac{1}{\left(1 - a\text{e}^{-j\omega}\right)^r} \end{array}$$

w/ restriction $$\left|a\right|<1$$ imposed on the latter; correct me if I am wrong but, from the definition of the DTFT...

$$X[\text{e}^{j\omega}] = \mathscr{F}[x[n]] = \sum_{n=-\infty}^{\infty}\left[x[n]\text{e}^{-j\omega n}\right]$$

The DTFT of $$u[n]$$ should be...

$$X[\text{e}^{j\omega}] = \mathscr{F}[u[n]] = \sum_{n=0}^{\infty}\left[\text{e}^{-j\omega n}\right] = \frac{\text{e}^{j\omega}}{-1 + \text{e}^{j\omega}}$$

Which should simplify to...

$$\Longrightarrow X[\text{e}^{j\omega}] = \frac{1}{1-\text{e}^{-j\omega}}$$

&, if my 2 steps were correct, the 1st DTFT pair should become a special case of the 2nd (when $$a = r = 1$$), but not w/o modifying the latter's restriction s.t. $$\left|a\right|\leq1$$.

The weird result leads me to believe that I made a mistake somewhere; could someone help me out?