1. Consider the ratio of its consecutive terms, $\left\lvert \frac{x^{n+1}}{n+1} \times \frac{n}{x^{n}} \right\rvert = \left\lvert \frac{n}{n+1}x \right\rvert \to |x|, \; \text{ as } n\to \infty $Therefore, by ratio test $\sum_{1}^{\infty} \frac{x^{n}}{n}$ converges for $|x|<1$ and diverges for $|x|>1$. When $x=1$, the series equals $\sum_{1}^{\infty} \frac{1}{n}$ which diverges since the partial sums are unbounded. When $x=-1$, the series equals $\sum_{1}^{\infty} \frac{(-1)^{n}}{n}$ which converges by alternating series test since $\frac{1}{n}$ converges to $0$, is decreasing and non-negative. 2. Suppose $|x|<1$ then $\sum_{n=1}^{\infty} |a_{n}x^{n}| \leq \sum_{n=1}^{\infty} x^{n}<\infty$. Therefore $\sum_{n=1}^{\infty}a_{n} x^{n}$ converges for $|x|<1$ since it converges absolutely. If $|x|>1$, $a_{n}x^{n} \not\to 0$ so the series diverges. 3. Follows from extreme value theorem since convergent power series is continuous. 4. Their graphs are shown below: ![[MA139 Assignment 1.png|400]] ![[MA139 Assignment 1 res1.png|400]]