For $z\in\mathbb{C}$ with real part greater than $1$ the sum $$\sum\_{p}{\frac{1}{p^z}},$$ where the sum is taken over all primes $p$, converges absolutely. It is also well known that the same sum with $z=1$ does not converge. Now my question is if there are $y\in\mathbb{R}$ such that
$$\sum\_{p}{\frac{1}{p^{1+iy}}}$$
converges? Many thanks in advance! 



Convergence of the summation 1/p^(1+iy)   (over all primes p with y a nonzero real number)
This is always convergent for any real $y \neq 0$. This follows from the fact that
the related integral 
$$\int\_2^\infty \frac{x^{iy-1}}{\log x} dx $$
is convergent (to see this use the substitution $t=\log x$ ), and say the prime number theorem with some weak error term, in fact
$$ \pi(x)=\frac x {\log x} \left( 1+O \left( \frac 1 {\log x} \right) \right) $$
is sufficient. In a similar spirit the similar sum
$$ \sum\_{n=1}^\infty n^{-1+iy}
$$ 
is not convergent for any $y$ (while bounded for $y \neq 0$ the partial sums will oscillate). This can be seen from the related integral
$$
 \int\_2^\infty x^{iy-1} dx 
$$ and the same substitution.


