DTFT of (1/2)^(n−1)u[n−1]

The signal $x[n] = \left(\tfrac{1}{2}\right)^{n-1}u[n-1]$ is the sequence $g[n] = \left(\tfrac{1}{2}\right)^n u[n]$ delayed by one sample: $x[n] = g[n-1]$.

Method 1: time-shift property

$$g[n] = \left(\tfrac{1}{2}\right)^n u[n] \leftrightarrow G(e^{j\omega}) = \sum_{n=0}^{\infty}\left(\tfrac{1}{2}e^{-j\omega}\right)^n = \frac{1}{1 - \frac{1}{2}e^{-j\omega}}$$

The geometric series converges because $|\tfrac{1}{2}e^{-j\omega}| = \tfrac{1}{2} < 1$. Then, by $g[n - n_0] \leftrightarrow e^{-j\omega n_0}G(e^{j\omega})$ with $n_0 = 1$:

$$\boxed{X(e^{j\omega}) = \frac{e^{-j\omega}}{1 - \frac{1}{2}e^{-j\omega}}}$$

Method 2: direct sum (check)

$$X(e^{j\omega}) = \sum_{n=1}^{\infty}\left(\tfrac{1}{2}\right)^{n-1}e^{-j\omega n} = e^{-j\omega}\sum_{m=0}^{\infty}\left(\tfrac{1}{2}e^{-j\omega}\right)^m = \frac{e^{-j\omega}}{1 - \frac{1}{2}e^{-j\omega}}$$

Here $m = n - 1$.

Magnitude: $|X(e^{j\omega})| = \frac{1}{\sqrt{1.25 - \cos\omega}}$. This is 2 at $\omega = 0$ and $2/3$ at $\omega = \pi$, the same as for $g[n]$. The delay only adds a phase of $-\omega$.