Uniform convergent of $\sum x^n/n$ It is said in Wikipedia that $\displaystyle \sum_{n\ge 1}\dfrac{x^n}{n}$ converges uniformly on $(-1,0)$ and converges absolutely at each point by the geometric series test. For the absolute convergence each point (Does this mean that it converges pointwise absolutely?) means that, for fixed $r\in(-1,0)$, $\displaystyle \sum_{n\ge 1} \left|\dfrac{r^n}{n}\right|\le \sum_{n\ge 1}|r|^n<\infty$ by geometric series. Am I right for this one?
I have a question for the uniform convergence. My attempt is that let $f_n(x)=x^n/n$ and the limit function is $f(x)=\log(1/(1-x))$ I guess?
For each $\epsilon>0$ we have
$$\left|\sum_{i\le n}f_i(x)-\log\left(\dfrac{1}{1-x}\right)\right|=\left|x+\dfrac{x^2}{2}+\cdots+\dfrac{x^n}{n}-\log\left(\dfrac{1}{1-x}\right)\right|=|x|^{n+1}\left|\dfrac{1}{n+1}+\dfrac{x}{n+2}+\cdots\right|.$$
The sum in the second absolute in the right-most term is $\displaystyle\le |1|+|x|/2+|x|^2/3+\cdots=\log(1/(1-|x|))$. How can I show that $\exists N$ such that for $n\ge N$, $|x|^{n+1}\log(1/(1-|x|))<\epsilon$? I am not sure even this way is valid or not?
It is also said that the uniform convergence (that is not absolute-uniform) may be rearrange the terms in the sum into the non-uniformly convergent? Could you give me examples for those? Thank you.
 A: The series $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$ converges, by the alternate series test. In other words, the power series$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n\tag1$$converges when $x=1$. It follows from Abel's test that the power series $(1)$ converges uniformly on $[0,1]$. If $x\in[0,1)$, it's a classical exercise on Taylor's theorem that the sum of the series is $\log(x+1)$; by continuity, it follows that the equality also holds when $x=1$. Now, replacing $x$ by $-x$ and multiplying everything by $-1$, we get that $\sum_{n=1}^\infty\frac{x^n}n$ converge uniformly to $-\log(1-x)$ on $[-1,0]$. But $-\log(1-x)=\log\left(\frac1{1-x}\right)$.
A: By integrating the geometric series with remainder, we have
$$
\log \left( {\frac{1}{{1 - x}}} \right) = \sum\limits_{k = 1}^n {\frac{{x^k }}{k}}  + \int_0^x {\frac{{t^n }}{{1 - t}}dt}  = \sum\limits_{k = 1}^n {\frac{{x^k }}{k}}  + x^{n + 1} \int_0^1 {\frac{{s^n }}{{1 - xs}}ds} .
$$
Now if $-1 \leq x \leq 0$, then
$$
\left| {x^{n + 1} \int_0^1 {\frac{{s^n }}{{1 - xs}}ds} } \right| \le \left| x \right|^{n + 1} \int_0^1 {s^n ds}  = \frac{{\left| x \right|^{n + 1} }}{{n + 1}} \le \frac{1}{{n + 1}}.
$$
This shows that the convergence is uniform on $-1 \leq x \leq 0$.
