= Solution
The <ratio test> states that, for a <series> $\sum a_n$ with nonzero terms eventually, if $|a_{n+1}/a_n|\to L<1$, the <series> converges absolutely; if $L>1$, including $L=\infty$, it diverges because its terms do not tend to zero. At $L=1$ the test is inconclusive: the <harmonic series> diverges while $\sum n^{-2}$ converges.
For $|x|<1$, the <ratio test> applied to $x^n/n$ gives <absolute convergence> when $x\ne0$, and at $x=0$ every such term vanishes. Thus
$$
\sum_{n=1}^N\frac{x^n-1}{n}
=\sum_{n=1}^N\frac{x^n}{n}-\sum_{n=1}^N\frac1n\longrightarrow-\infty.
$$
At $x=1$ all terms are zero, so the <series> converges to zero. At $x=-1$, the even terms vanish and the odd terms are $-2/n$, whose <partial sums> tend to $-\infty$. Finally, if $|x|>1$, then $|x^n-1|/n\geq(|x|^n-1)/n\to\infty$, so the <term test for divergence> applies. Consequently
$$
\boxed{\text{The series converges exactly for }x=1,\text{ and its sum there is }0.}
$$
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