I’m having trouble with an inequality in my book, namely, if $P(z)=a_o+a_1z+a_2z^2+...+a_nz^n$ where $a_i\in\mathbb{C}$, $a_n\neq0$, and $n\in\mathbb{Z}^+$ then: $$\left|\frac{1}{P(z)}\right|<\frac{2}{|a_n|R^n}\space\space \text{ when $|z|>R$}$$
which is implied by the inequality in the title.
It is proved in the book as follows:
Let $$w=\frac{a_0}{z^n}+\frac{a_1}{z^{n-1}}+\frac{a_2}{z^{n-2}}+...+\frac{a_{n-1}}{z} \space \space \space \text{$(z\neq0$)}$$
so that $P(z)=(a_n+w)z^n$ when $z\neq0$. Thus $wz^n=a_o+a_1z+a_2z^2+...+a_{n-1}z^{n-1}$ which tells us by the Triangle Inequality: $$|w||z|^n\leq|a_o|+|a_1||z|+|a_2||z|^2+...+|a_{n-1}||z|^{n-1}$$ $\iff$ $$|w|\leq\frac{|a_0|}{|z|^n}+\frac{|a_1|}{|z|^{n-1}}+\frac{|a_2|}{|z|^{n-2}}+...+\frac{|a_{n-1}|}{|z|}$$
Now the portion I don’t understand:
Note a sufficiently large positive number $R$ can be found such that each of the quotients on the right in the inequality above is less than the number $\frac{|a_n|}{2n}$ when $|z|>R$, and so...
The rest of the proof makes sense to me, but I’ll reproduce it below for convenience
$$|w|<n\frac{|a_n|}{2n}=\frac{|a_n|}{2}\space\space\space\text{when $|z|>R$}$$
Therefore since $|w|<\frac{|a_n|}{2}$ $$|a_n+w|\geq ||a_n|-|w||=\left|\frac{2|a_n|}{2}-|w|\right|> \left|\frac{2|a_n|}{2}-\frac{|a_n|}{2}\right|=\frac{|a_n|}{2} \space\space\space\text{when $|z|>R$} $$
Then since $P(z)=(a_n+w)z^n$, we have $$|P(z)|=|a_n+w||z|^n>\frac{|a_n|}{2}|z|^n> \frac{|a_n|}{2}R^n \space\space \text{ when $|z|>R$} $$
which $\iff \left|\frac{1}{P(z)}\right|<\frac{2}{|a_n|R^n}\space\space \text{ when $|z|>R$}$.
My question is: How and why do we choose such an $R$ to achieve the bound $\frac{|a_n|}{2n}$ in particular?
Why not $\frac{|a_n|}{3n}$ or $\frac{|a_n|}{2n^2}$, etc? Is it just because $\frac{|a_n|}{2n}$ makes the inequality “work?”
Then what about the constant factor of $2$? That seemed arbitrary as well.
Any explanation here would be awesome. Thank you so much!