Let $E$ be the expected number of moves, and $E_x$ be the expected number of moves the ant takes when it has already visited $x$ vertices to visit a new vertex.
Using linearity of expectation,
$$E = \sum^{n-1}_{i=1} E_i$$
$E_i$ is, as you have suggested, the expected value of a geometric distribution with $p=\frac{n-i}{n-1}$.
$$\implies E = \sum^{n-1}_{i=1} \frac{n-1}{n-i}$$
$$= (n-1)\sum^{n-1}_{i=1} \frac{1}{n-i}$$
$$= (n-1)\sum^{n-1}_{i=1} \frac{1}{i}$$
$$= (n-1)H_{n-1}$$
where $H_n$ is the nth harmonic number