# 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.

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 $\;1+\epsilon <2\;$ ...

• In a comment just prior to the above theorem, Edwards states that [something] "is therefore a consequence of the case $\epsilon = 1$ of the ...[theorem]. That makes me think that epsilon is not limited to < 1. Commented Jan 10, 2018 at 0:24
• On further reflection, I think you have it exactly right. Edwards proof is only intended for the case $\epsilon < 1$. (The $16$ in the inequality is the giveaway). That said, I believe the Theorem is intended for ALL $\epsilon > 0$. Now I speculate. Perhaps Edwards thought the case where $\epsilon \geq 1$ is simply obvious. Once you prove for $\epsilon < 1$, the comparison test gives you $\epsilon \geq 1$ because for all but a finite number of ordered roots, we have $|p - 1/2| > 1$. Commented Jan 10, 2018 at 16:40

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

$$0 \leqslant \lim_{n \to \infty} \frac{n^4}{\exp(n^a)} \leqslant \lim_{n \to \infty} \frac{m!n^4}{n^{am}} = 0.$$