Riemann Zeta Function (Edwards), Section 2.5 - Help with Proof Background.  I am working my way through the book Riemann's Zeta Function (Edwards), and am stuck on one line of the proof in Section 2.5. The theorem/proof is short enough that I show it in full below.  I pose my question immediately following the proof.
Theorem. For any given $\epsilon > 0$, the series $\displaystyle \sum_p \frac{1}{|(p- 1/2)|^{1 + \epsilon}}$ converges, where $p$  ranges over all roots $p$ of  $\xi(p)=0$. (NOTE: $\xi(p)$ is Riemann's completed zeta function).
Proof. Let the roots $p$ be numbered $p_1, p_2, ...$ in order of increasing $|p - 1/2|$.   Let $R_1, R_2, ...$  be the sequence of positive real numbers defined implicitly by the equation  $4R_n \log(R_n) = n$.  Then [by a prior Edwards proof] there are at most $3n/4$ roots $p$ inside the circle $|s - 1/2| = R_n$. Therefore, the $n$th root is not in the circle; that is, 
$|p_n - 1/2| > R_n$. Thus:
\begin{align*}
\sum_p \frac{1}{|(p_n - 1/2)|^{1 + \epsilon}} &\leq \sum_p \frac{1}{(R_n)^{1 + \epsilon}} = \sum_p \frac{(4 \cdot \log(R_n))^{1 + \epsilon}}{n^{1 + \epsilon}} \\
&=\sum_p \frac{1}{n^{1 + (\epsilon/2)}} \cdot \frac{(4 \cdot \log(R_n))^{1 + \epsilon}}{n^{(\epsilon/2)}}
\end{align*}
Now $\log(n) = \log(R_n) + \log(4) + \log(\log(R_n)) > \log(R_n)$ for $n$ large.  Hence, $(4 \cdot \log(R_n))^{1 + \epsilon} < 16 \cdot (\log(n))^2 < n^{\epsilon/2}$ for all sufficiently large $n$ and:
\begin{align*}
\sum_p \frac{1}{|(p_n - 1/2)|^{1 + \epsilon}} < \text{const} + \sum_{n=1}^{\infty} \frac{1}{n^{1 + (\epsilon/2)}} < \infty
\end{align*}
as was to be shown.
Question. I cannot come up with a proof for the statement that $(4 \cdot \log(R_n))^{1 + \epsilon} < 16 \cdot (\log(n))^2$ for sufficiently large $n$.  I have tried L'Hospital.  I have tried substitutions for $n$, for $R_n$, and for $\log(R_n)$, but have made no progress.  Any assistance would be appreciated.
 A: An idea with an assumption (an easy and "clear" one, I think): 
Observe that from the line above of what you can't prove, we have
$$\log R_n<\log n\implies 4\log R_n<4\log n\implies \left(4\log R_n\right)^{1+\epsilon}<(4\log n)^{1+\epsilon}<16\log^2n$$
as I assume $\;\epsilon>0\;$ is rather small and thus $\;1+\epsilon <2\;$ ...
A: To prove the theorem for all $\epsilon > 0$, the (dubious) intermediate step is not needed. It is enough to show without this step that for sufficiently large n,
$$\tag{*}(4 \log R_n)^{1 + \epsilon} < n ^{\epsilon/2}.$$
Since $\log R_n < \log n$ for large $n$, the inequality (*) follows if 
$$(4 \log n)^{1 + \epsilon} < n ^{\epsilon/2}$$
or
$$n^4 < \exp(n^a)$$
where $a = \epsilon/(2 + 2\epsilon).$
This, of course, is true for any $a > 0$ and sufficiently large $n$ since  $\exp(n^a) = \sum_{k=0}^\infty n^{ak}/k!$. Choosing an integer $m$ such that $am > 4$, we have 
$$0 \leqslant \lim_{n \to \infty} \frac{n^4}{\exp(n^a)} \leqslant \lim_{n \to \infty} \frac{m!n^4}{n^{am}} = 0. $$
