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!
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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.
