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 $\lvert z \rvert > R$
 A: 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 $$|z|>R_0>\max\left\{\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}},\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}},...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}},\frac{2n|a_{n-1}|}{|a_n|}\right\}$$$$\implies|z|>\left(\frac{2n|a_k|}{|a_n|}\right)^{\frac{1}{n-k}}\text{ for each }k=0,1,...,(n-2),(n-1)$$$$\implies |z|^{n-k}>\frac{2n|a_k|}{|a_n|}\text{ for each }k=0,1,...,(n-2),(n-1)$$$$\implies\frac{|a_n|}{2n}>\frac{|a_k|}{|z|^{n-k}}=\frac{|a_k|}{|z^{n-k}|}=\left|\frac{a_k}{z^{n-k}}\right|\text{ for each }k=0,1,...,(n-2),(n-1).$$
A: Because if we take $\epsilon>0$, then for large enough $R$ all of
$|a_0|/R^n,|a_1|/R^{n-1},\ldots,|a_{n-1}|/R$ are $<\epsilon$.
A: Take the inequality ||> as true. Both are scalars and real, then according to the theorem the reals are ordered: 1/||<1/.
Since the potential is preserving the order for all positive integer n: 1/z^n<^/^n. Consider the numbers from 1 to n then this inequality is true for all these numbers.
Now
=()/z^n.
We know not much about the sequence a_i i=1,...n-1, but if  goes larger and larger the well ordered of the reals warrants that each summand will get smaller than 1. But that is a fairly upper bound. The largest term for z->0 is the last term in   because it is only -1 in the exponent. It does not matter how small or
big −1 is, for large enough values of z it starts to dominate the sum.
A simple example is this:

This is the case all coefficients equal and n=3. The simple sum of the coefficients is an upper limit for the  over the reals.
Since the b=Max[{a(m)/z^m:0<=m<=n-1}] has a  such that b/z^n-k<a(n-1)/z k the index of the maximum of the coefficients.
It is possible to replace the summands with this maximum: (n-1)b/z^n-k<n-1)a(n-1)/z

The green line is the  |()|/2 in the choosen example.
At this point of the discussion, it is necessary that all a(m)>0 if 0 then this summand is irrelevant to the discussion.
The larger the a(m) are the larger  but the point is warranted by the well order of the reals.
If n the number of coefficients is fixed, but z raises to infinity z always dominates and the sum gets smaller, the first derivative in negative on the positive reals.
< ||/2
independent of the magnitude of a(n). There can too be another arbitrary positive number or even smaller ones be selected and this occurs for larger and larger z.
The point is that  can be greater than n max[{a(m),0<=m<=n-1}]/z and this is
steadily decreasing for z getting bigger. a(n)/(2n) is another limit that is beaten be increasing z. Therefore exists  for sure. |()|/2 is a constant for the problem during z and  can be chosen arbitrarily large.
lim w for z->oo is 0!
A: Statement
Sometimes, it helps to zoom out and look at an even more general setting. Basically, this is what you want to prove.

Lemma: For any complex number $b$, for any positive real number $\varepsilon$, there exists a positive real $R$ such that, for every complex $z$ and every positive integer $n$,
$$
    |z|>R \hspace{2mm}\implies\hspace{2mm} \left| \frac{b}{z^n} \right| < \varepsilon.
$$

Why is this the thing that you want to prove? Here is why. After we prove this lemma is true, we have the following corollary.

Corollary: For any finite quantity of complex numbers $a_0, \ldots, a_{n-1}$, for any positive real number $\varepsilon$, there exists a positive real $R$ such that, for every complex $z$,
$$
    j\in\{1,\ldots,n\}, \hspace{1mm} |z|>R \hspace{2mm}\implies\hspace{2mm} \left| \frac{a_{n-j-1}}{z^j} \right| < \varepsilon.
$$

One way to see that this corollary follows from the preceding lemma is like this:
For every $j\in\{0,\ldots,n-1\}$, our lemma gives us an $R_j$ such that $|z| > R_j \implies \left| \frac{a_{n-j-1}}{z^j} \right| < \varepsilon$. Thus, since we only have a finite quantity of terms to worry about, we can choose $R = \max\{R_0,\ldots,R_{n-1}\}$.
Lastly, of course, your problem is solved by choosing $\varepsilon = \frac{|a_n|}{2n}$ (recall, $n$ is not "going to infinity" so this is a plain old real number like any other).

Proof of Lemma
Let $b$ be any complex number. Let $n$ be a positive integer. Then,
$$
     \left| \frac{b}{z^n} \right|
=
     \frac{|b|}{|z|^n}.
$$
We are supposed to be able to make this smaller that $\varepsilon$. Our strategy will be to force the denominator to be big.
More precisely, give me an arbitrary $\varepsilon>0$. I will show that a real number $R$ can be chosen in a way which doesn't depend on $n$ so that if $|z|>R$ then $\frac{|b|}{|z|^n} < \varepsilon$. How do I know that such an $R$ can be chosen? Why, because here it is! Choose
$$
    R = \max\left\{ 1, \frac{|b|}{\varepsilon} \right\}.
$$
Now, give me any complex number $z$ such that $|z|>R$. I will verify that we indeed have the desired outcome. Using our choice of $R$, the statement $|z|>R$ means precisely that two things are true:
$$
    |z| > \frac{|b|}{\varepsilon}
\hspace{4.5mm}\text{and}\hspace{4mm}
    |z| > 1.
$$
Since $|z|$ is greater than 1, we can multiply an inequality by $|z|$ on both sides and preserve the order of the inequality. Therefore
$$
    |z| > 1
\hspace{4mm}\text{leads to}\hspace{4mm}
    |z|^2 > |z|.
$$
Of course, we can do this as much times as we like, to conclude that
$$
    \cdots > |z|^3 > |z|^2 > |z|.
$$
More precisely, it is true that, since $|z|>1$, we have $|z|^n \geq |z|$ for every positive integer $n$, as you can verify by induction if you like.
Now, since $|z|^n \geq |z|$ for every $n\geq1$ and $|z| > \frac{|b|}{\varepsilon}$, we can slap these two inequalities together to obtain that
$$
    |z|^n > \frac{|b|}{\varepsilon} \text{ for every $n>1$}.
$$
Last, we can take that inequality and multiply it by $\frac{\varepsilon}{|z|^n}$ on both sides (notice that $|z|$ is not zero since $|z|>1$). Doing so, and cancelling, we obtain that
$$
     \varepsilon > \frac{|b|}{|z|^n} \text{ for every $n>1$}.
$$
End of proof.

Is this answering?
