the inequaltiy in this picture
I try to prove it but i stuck with $\frac2{|a_n|}$. How did he got it ??
I followed the proof in the textbook but the term $\frac{2}{|a_n|}$ does not make sense to me.
It said $$ |w| \leq \sum_{i=0}^{n-1} \frac{|a_i|}{|z^{n-i}|}.$$ As $R$ increases, we have $$|w| < n\frac{|a_n|}{2n}$$ and this step which i did not get it
As $|z| \rightarrow \infty $, $\frac{|a_i|}{|z|^{n-i}} \rightarrow 0$ for $0\leq i<n$
That is when $|z|$ is sufficiently large, we can make $\frac{|a_i|}{|z|^{n-i}}$ arbitrarily small in magnitude.
That is $\forall \epsilon > 0$, $\exists R_{\epsilon, i} > 0$, $|z| > R_{\epsilon,i} \implies \frac{|a_i|}{|z|^{n-i}}<\epsilon.$
Since $a_n \neq 0$, we can let $\epsilon = \frac{|a_n|}{2n}>0$ and we know that
$\exists R_{ i} > 0$, $|z| > R_{i} \implies \frac{|a_i|}{|z|^{n-i}}<\frac{|a_n|}{2n}.$
Pick the same $R$ for all the $i \in \{ 0, \ldots , n-1 \}$, for example, let $R= \max \{ R_0, \ldots, R_{n-1}\}$.
hence if $|z|>R$, $$|w|\leq \sum_{i=0}^{n-1}\frac{|a_i|}{|z|^{n-i}} < \sum_{i=0}^{n-1}\frac{|a_n|}{2n} = n \frac{|a_n|}{2n}=\frac{|a_n|}{2}$$