Why is $\frac{a_k}{z^{n-k}},(0\leq k < n)$ less than $\frac{\lvert a_n \rvert}{2n}$ Suppose there is a polynomial:
$$P(z) = a_0 + a_1z +a_2z^2 +\dots +a_nz^n,\quad (a_n \ne 0)$$
Let $$w = \frac{a_0}{z^n} + \frac{a_1}{z^{n-1}} + \frac{a_2}{z^{n-2}} + \dots + \frac{a_{n-1}}{z},$$
Why is it that for a sufficiently large positive number $R$, the modulus of each of the quotients in the expression $w$ is less than the number $\frac{\lvert a_n \rvert}{2n}$ when $\lvert z \rvert > R$

Since, $a_n\not=0$ we have $|a_n|\not=0$. Now, we have to find a positive real number $R_0$  such that for any complex number $z$ with $|z|> R_0$ we have $$\left |\frac{a_0}{z^n}\right|=\frac{|a_0|}{|z^n|}=\frac{|a_0|}{|z|^n}<\frac{|a_n|}{2n},$$$$\left |\frac{a_1}{z^{n-1}}\right|=\frac{|a_1|}{|z^{n-1}|}=\frac{|a_1|}{|z|^{n-1}}<\frac{|a_n|}{2n},$$$$\vdots$$$$\left |\frac{a_{n-2}}{z^2}\right|=\frac{|a_{n-2}|}{|z^2|}=\frac{|a_{n-2}|}{|z|^2}<\frac{|a_n|}{2n},$$$$\left |\frac{a_{n-1}}{z}\right|=\frac{|a_{n-1}|}{|z|}<\frac{|a_n|}{2n}.$$
In other words,  we have to find a positive real number $R_0$  such that for any complex number $z$ with $|z|> R_0$ we have $$\frac{2n|a_0|}{|a_n|}<|z|^n,\frac{2n|a_1|}{|a_n|}<|z|^{n-1},...,\frac{2n|a_{n-2}|}{|a_n|}<|z|^2,\frac{2n|a_{n-1}|}{|a_n|}<|z|.$$ But, the above inequalities hold if and only if $$\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}}<|z|,\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}}<|z|,...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}}<|z|,\frac{2n|a_{n-1}|}{|a_n|}<|z|.$$ So, we may take $$R_0:=1+\max\left\{\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}},\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}},...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}},\frac{2n|a_{n-1}|}{|a_n|}\right\}.$$ So, for any complex number $z$ with $$|z|>R_0>\max\left\{\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}},\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}},...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}},\frac{2n|a_{n-1}|}{|a_n|}\right\}$$$$\implies|z|>\left(\frac{2n|a_k|}{|a_n|}\right)^{\frac{1}{n-k}}\text{ for each }k=0,1,...,(n-2),(n-1)$$$$\implies |z|^{n-k}>\frac{2n|a_k|}{|a_n|}\text{ for each }k=0,1,...,(n-2),(n-1)$$$$\implies\frac{|a_n|}{2n}>\frac{|a_k|}{|z|^{n-k}}=\frac{|a_k|}{|z^{n-k}|}=\left|\frac{a_k}{z^{n-k}}\right|\text{ for each }k=0,1,...,(n-2),(n-1).$$
