Uniform Convergence of $\sum_{n=2}^\infty z^n/(n\ln(n))$ over $\overline{B(0,1)}\backslash B(1,\varepsilon)$ Let $\varepsilon >0$, show the uniform Convergence of $\sum_{n\ge2} \frac{z^n}{n\ln(n)}$ over $\overline{B(0,1)}\backslash B(1,\varepsilon)$.
I first found that the radius $R=1$. Next, I noticed that $\sum_{n\ge2}\frac{1}{n\ln(n)}$ diverges and $\sum_{n\ge2}\frac{(-1)^n}{n\ln(n)}$ converges.
However, after that, I am stuck. I know that $\forall r<1, \sum_{n\ge2} \frac{z^n}{n\ln(n)}$ converges uniformly over $\overline {B(0,r)}$ because it is a power series.
As a hint, I need to use the summation by parts.
Just in case : $B(a,r) = \left\{z\in\mathbb C\,\middle|\, |a-z| < r\right\}$. I appreciate any feedback !
 A: You only use that $(a_n)_{n \ge 2} := 1/(n \ln(n))$ decreases and converge to $0$. You may use Abel transformation to prove that uniform Cauchy criterion holds. Set
$$S_n(z) = \sum_{k=2}^n a_kz^k \text{ and } T_n(z) = \sum_{k=n}^{+\infty} z^k = \frac{z^n}{1-z}.$$
For every $q > p \ge 2$ and $z \in \overline{D}(0,1) \setminus D(1,\epsilon)$, one has
$$S_q(z)-S_p(z) = \sum_{k=p+1}^q a_nz^n = \sum_{k=p+1}^q a_n T_n(z) - \sum_{k=p+1}^q a_n T_{n+1}(z).$$
$$S_q(z)-S_p(z) = \sum_{k=p+1}^{q} a_n T_n(z) - \sum_{k=p+2}^{q+1} a_{n-1} T_{n}(z).$$
$$S_q(z)-S_p(z) = \sum_{k=p+1}^q (a_n-a_{n-1})T_{n}(z) + a_{p}T_{p+1}(z) - a_{q}T_{q+1}(z).$$
$$|S_q(z)-S_p(z)| \le \sum_{k=p+1}^q (a_{n-1}-a_n)|T_{n}(z)| + a_p|T_{p+1}(z)| + a_{q}|T_{q+1}(z)|.$$
Since for every $k \ge 0$
$$|T_k(z)| \le \frac{1}{|1-z|} \le \frac{1}{
\epsilon},$$
we derive
$$|S_q(z)-S_p(z)| \le \frac{1}{\epsilon}\Big(\sum_{k=p+1}^q (a_{n-1}-a_n) + a_p+a_q \Big) = \frac{2a_p}{\epsilon}.$$
Since $a_p \to 0$ as $p \to +\infty$, uniform Cauchy criterion applies.
