I need to prove that if $f$ is analytic in the unit disk and $|f(z)|\leq \frac{1}{1-|z|}$ for z all $z\in D_1(0)$ then $|f'(0)|\leq 4$. This is my proof and I need to verify this.
Let $n\in \mathbb{N}$. Then $\overline{D}_{1-(\frac{1}{n+1})}(0) \subset D_1(0)$. Thus $f$ is analytic in the closed disk $\overline{D}_{1-(\frac{1}{n+1})}(0)$. Then by Cauchy's estimate $$|f'(0)|\leq \frac{1}{1-\frac{1}{n+1}} \frac{1}{1-|1-\frac{1}{n+1}|}=\frac{(n+1)^2}{n}$$. This is where I am sort of stuck. Since the radius of $\overline{D}_{1-(\frac{1}{n+1})}(0)$ is minimum when $n=1$ substituting this to the above I can get that $$|f'(0)|\leq4$$. How can I improve this? Thanks