Is power series $\sum_{n=1}^\infty\frac{x^n}{n}$ uniformly convergent in $(-1,1)$

Question

I know that if $\sum a_nx^n$ has radius of convergence R then this power series converges uniformly for every compact subset of $(-R,R)$ and now no doubt converges pointwise in $(-R,R)$ but little bit confused what can we say about uniform convergence in $(-R,R)$.

As for example what can we say about uniform convergence of series $\sum_{n=1}^\infty\frac{x^n}{n}$ in (-1,1)?

Answer 1

Hint: Any partial sum is bounded, but the series converges pointwise to something unbounded.

Answer 2

Here's a mostly painless way to show unboundedness, in order to apply Arthur's hint.

As you note $s = \sum_j x^j/j$ converges over compact sets, and moreover so does the series of term-wise derivatives $\sum_{j \geq 1}x^{j-1} = \sum_{j \geq 0}x^j = 1/(1-x)$. A standard theorem of series gives $s' = 1/(1-x)$ and so integrating and using that $s(0) = 0$ one gets

$$ s(x) = -\mathbf{ln}(1-x) $$

and then $s(x) \xrightarrow{x \to _+1} +\infty$.

Answer 3

If it is uniformly convergent then there exists $m$ such that $ |\sum\limits_{k=i}^{j}\frac {x^{k}} k| <1$ for all $x$ for all $j>i>m$. Let $x \to 1$ in this. Do you see a contradiction?