Discuss the convergence and uniform convergence on $[0,1)$
$$ \sum_{n=1}^{\infty}{\frac{x^n}{n}}$$
For example , for the question how can I prove $f_n = \frac{x^n}{n}$ convergence pointwise to a certain function $f$ here is what I did so far , I fixed $x \in [0,1) $ but I’m having trouble computing $\lim_{n \rightarrow \infty} f_n$
And for the uniform convergence I tried to find the maximum of $ | f_n - f | $ but it doesn’t go anywhere .
Thanks in advance.
For uniform convergence it is necessary that $\sup_{x \in [0,1)}|S_n(x) - S(x)| \to 0$ as $n \to \infty$ where $(S_n)$ is the sequence of partial sums. You don't really need to know what $S(x)$ is to proceed.
For any $x \in [0,1)$ we have
$$\sup_{x \in [0,1)} |S_n(x) - S(x)| = \sup_{x \in [0,1)} \sum_{k=n+1}^\infty \frac{x^k}{k} \geqslant \sum_{k=n+1}^{2n} \frac{x^k}{k}\geqslant n \cdot \frac{x^{2n}}{2n} = \frac{x^{2n}}{2} $$
Thus, with $x = 1-1/n$,
$$\sup_{x \in [0,1)} \sum_{k=n+1}^\infty \frac{x^k}{k} \geqslant \frac{(1- 1/n)^{2n}}{2} \underset{n \to \infty}\longrightarrow \frac{e^{-2}}{2} \neq 0,$$
which shows that convergence of the series is not uniform.