Reading a textbook on basic complex analysis I stumbled upon the following result: The series $\sum_{n=1}^{\infty}\tfrac{z^n}{n}$ converges uniformly and absolutely in the closed disk $\overline{D(0,r)}$ but does not converge uniformly on $D(0,1)$.
My question comes with the argument that is used to show that the series indeed does not converge uniformly on $D(0,r)$ which is as follows:
Say it does converge uniformly, then, using Cauchy criteria for uniform convergence, we have that for every $\epsilon>0$ exists $N\in\mathbb{N}$ s.t. if $n>N$, $$ \frac{x^n}{n}+\frac{x^{n+1}}{n+1}+\cdots+\frac{x^{n+p}}{n+p}<\epsilon,\quad\forall x\in[0,1),\quad\forall p\in\mathbb{N}. $$ Now, since the harmonic series diverges, using again Cauchy criteria, we have that $$ \frac{1}{N}+\frac{1}{N+1}+\cdots+\frac{1}{N+p}>2\epsilon.\tag1 $$ Then we have $$ \frac{x^N}{N}+\frac{x^{N+1}}{N+1}+\cdots+\frac{x^{N+p}}{N+p}>x^{N+p}\left(\frac{1}{N}+\frac{1}{N+1}+\cdots+\frac{1}{N+p}\right)>2\epsilon\cdotp x^{N+p}. $$ Now, using the intermediate value theorem we have that there exists $x\in[0,1)$ s.t. $x^{N+p}>1/2$, and for that $x$ we have a contradiction with $(1)$.
And now to my question, the same argument holds for any $r$ s.t. $r>1/2$? Then we would have that the series does not converge in any closed disk $\overline{D(0,r)}$ with $r>1/2$.
Seems to me that I am missing something.
First of all we need to prove it is not uniformly convergent on D(0,1) and then $$ \frac{x^N}{N}+\frac{x^{N+1}}{N+1}+\cdots+\frac{x^{N+p}}{N+p}>x^{N+p}\left(\frac{1}{N}+\frac{1}{N}+\cdots+\frac{1}{N}\right)>2\epsilon\cdotp x^{N+p}. $$ This may not be true...
I suggest you to watch https://www.youtube.com/watch?v=bYfRWN8mhg4&list=PLDzvuf9Uf4FPFRWaMmBWNtbUukooikwBc&index=4