Convergence of a modified sum of prime reciprocals for all $s \in \mathbb{C}$? It is known that $\displaystyle \sum^\infty_{p \in \mathbb{P}} \frac{1}{p^s}$, with $\mathbb{P}$ the set of primes, only converges for $\Re(s) > 1$. 
The following sum of primes seems to converge for all $s \in \mathbb{C}$, with a (trivial) single pole at $-1$: 
$$\displaystyle \sum^\infty_{p \in \mathbb{P}} \left( \frac{1}{p-\frac{1}{p^s}}- \frac{1}{p+\frac{1}{p^s}} \right)$$
I guess that when $p \rightarrow \infty$, the incremental sum term $\rightarrow 0$, hence the infinite sum will converge (assuming $s$ is not infinitely close to $-1$). Is this reasoning correct?
Curious to learn if this infinite sum has any zeros in the complex domain (for $s \in \mathbb{R}$ it doesn't).
 A: 
I guess that when $p\to\infty$, the incremental sum term $\to 0$, hence the infinite sum will converge (assuming $s$ is not infinitely close to $−1$). Is this reasoning correct?

That the term of the series converges to $0$ is necessary, but not sufficient for the series to converge. We need to take a closer look at the terms.
$$\begin{align}
\frac{1}{p-\frac{1}{p^s}} - \frac{1}{p+\frac{1}{p^s}}
&= \frac{\left(p + \frac{1}{p^s}\right) - \left(p-\frac{1}{p^s}\right)}{p^2 - \frac{1}{p^{2s}}}\\
&= \frac{2}{p^s\left(p^2 - p^{-2s}\right)}\\
&= \frac{2}{p(p^{1+s} - p^{-(1+s)})}.
\end{align}$$
That shows that for $\operatorname{Re} s \neq -1$ the series converges absolutely, since
$$\lvert p(p^{1+s}-p^{-(1+s)})\rvert = p^{1+\lvert\operatorname{Re} 1+s\rvert}\cdot \left\lvert 1- p^{-2\lvert \operatorname{Re} 1+s\rvert}\right\rvert.$$
For $s = -1+it$ with $t\in \mathbb{R}\setminus \{0\}$, we have
$$p^{it} - p^{-it} = 2i\sin (t\log p),$$
and the term becomes
$$\frac{1}{ip\sin(t\log p)},$$
so the series certainly doesn't converge when $t = \frac{k\pi}{\log p}$ for some $k\in\mathbb{Z}\setminus\{0\}$ and some prime $p$. Since $\log p$ becomes arbitrarily large, these points are dense on the line $\operatorname{Re} s = -1$. I would expect that the series converges for no $s$ with $\operatorname{Re} s = -1$, but I don't see a proof.
