I feel sort of shaky over my attempt at this proof and would really appreciate someone else providing some insight.
Problem: $f(x)=\sum_{n=1}^\infty \frac{x^n}{n}$ converges for $x \in [-1,1)$. For a fixed $x_0\in(-1,1)$, use the Weierstrass M-Test to show $f(x)$ is continuous at $x_0$.
Attempt: Note that since the interval $(0,1)$ is open, there exists an $\epsilon >0$ such that $V_\epsilon (x_0) \subseteq (0,1)$. Consider $x_0+\epsilon$ and let $M_n=\frac{(x_0+\epsilon)^n}{n}$. Then we have $$\left| \frac{x^n}{n} \right| \leq M_n$$ for all $x \in (-x_0-\epsilon, x_0+\epsilon)$. Since $\sum_{n=1}^\infty M_n$ converges, then by the Weierstrass M-Test $\sum_{n=1}^\infty \frac{x^n}{n}$ converges uniformly on $(-x_0-\epsilon, x_0+\epsilon)$ to some function $f$. And by the Term-by-term Continuity Theorem, since each $f_n$ is continuous (all polynomials), and $\sum_{n=1}^\infty f_n$ converges uniformly, $f$ is continuous on $(-x_0-\epsilon, x_0+\epsilon)$ and hence $x_0$.
This answer might come a little late for you, but I tackled this exercise today. The strategy is to compare the series of functions with a geometric series.
Proof:
For all $n$, we have $\left\lvert \frac{x^{n}}{n} \right\rvert < x^{n}$. The geometric series $\sum_{n=0}^{\infty}ax^{n}$ with $\lvert x \rvert < 1$ converges. In our case, $a=1$. By the Weierstrass M-test, $f(x)$ converges uniformly.
To prove continuity, we combine the uniform convergence result with the fact that $f_{n}(x) = \frac{x^{n}}{n}$ is continuous for all $n$ on all reals since it is a polynomial function. Then, we have that $f(x)$ is continuous by the term-by-term continuity theorem.