Convergence of sum in proof that $\Phi(s) - \frac{1}{s-1}$ extends to $\Re(s) > \frac{1}{2}$

Question

Definitions: $\Phi(s) = \displaystyle\sum_{p} \frac{\log p}{p^s}$ where $p$ denotes a prime number.

$\zeta(s) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}$ denotes the Riemann zeta function.

---

I am proving this lemma:

Lemma: The function $\Phi(s) - \frac{1}{s-1}$, initially defined for $\Re(s) > 1$, extends to a meromorphic function on the half-plane $\Re(s) > \frac{1}{2}$, and is analytic for $\Re(s) \geq 1$.

In the proof, we derive the following relationship:

$$- \frac{\zeta'(s)}{\zeta(s)} = \Phi(s) + \displaystyle\sum_p \frac{\log p}{p^s(p^s - 1)}.$$

The proof then goes on to say "The infinite sum converges to an analytic function for $\Re(s) > \frac{1}{2}$ by comparison with $\displaystyle\sum_{n=1}^{\infty} \frac{\log(n)}{n^{2s}}$."

This is where I get stuck. Here is why:

I start by bounding the sum by absolute value, then throwing in the rest of the terms that have been omitted by summing over primes, then pull the absolute value inside:

$$\begin{array}{ll} \left| \displaystyle\sum_p \frac{\log(p)}{p^s (p^s - 1)} \right| &\leq \left| \displaystyle\sum_{n \in \mathbb{Z}^+} \frac{\log(n)}{n^s (n^s - 1)} \right| \\ &\leq \displaystyle\sum_{n \in \mathbb{Z}^+} \frac{|\log(n)|}{n^{\Re(s)} (n^{\Re(s)} - 1)} \\ &= \displaystyle\sum_{n \in \mathbb{Z}^+} \frac{|\log(n)|}{n^{2\Re(s)} - n^{\Re(s)}}. \end{array}$$

If the quotation above from the paper is to be believed, the next step should look like

$$\leq \displaystyle\sum_{n \in \mathbb{Z}^+} \frac{|\log(n)|}{n^{2\Re(s)}},$$

which can be shown convergent by the integral test. But to do this would require that

$$\frac{1}{n^{2 \Re(s)} - n^{\Re(s)}} \leq \frac{1}{n^{2\Re(s)}},$$

but I have a problem with that statement because it would imply

$$n^{2 \Re(s)} - n^{\Re(s)} \geq n^{2\Re(s)},$$

which is not true.

Am I missing something or is there an error in the paper? If it is an error, how should I proceed to show that the series converges?

Answer 1

This follows from the limit comparison test (instead of the plain comparison test): $$\lim_{n\rightarrow\infty}{n^{2\sigma}\over n^\sigma(n^\sigma-1)}=1$$ So either both $\sum_{n\ge 2}{\log(n)\over n^\sigma(n^\sigma-1)}$ and $\sum_{n\ge 2}{\log(n)\over n^{2\sigma}}$ converge or diverge.

Answer 2

Let me write $\sigma$ for the real part of $s$. Perhaps you can prove that for $\sigma\gt1/2$ and $n\ge2$ you have something like $n^{2\sigma}-n^{\sigma}\ge(1/1000)n^{2\sigma}$, which would be good enough.