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By using the ratio test, I'm able to show that $\sum \frac{x^n}{n}$ converges when $x\in (-1, 1).$

I'm trying to prove the same fact using the M-Test but can't see a suitable $M_n \geq \frac{x^n}{n}$ (with $M_n$ independent of $x$) to use. Of course, I can't use $\frac{1}{n}$, but that's all I can see. What would work here?

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    Since the series doesn't converge uniformly on $(-1,1)$, there are no $M_n$. On every subinterval $[-q,q]$, where $0 < q < 1$, you can take $q^n/n$. – 2017-01-16
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    @DanielFischer I see, that makes sense. – 2017-01-16
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    A remark: when it is convergent, the sum of this series is $-ln(1-x)$. – 2017-01-16

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For any $b<1$ we have for $x

hence the series converges on $ [-b,b]$. For $x = -1$ it converges as it is an alternating series.