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Is the sum $\sum_{\rho} \frac {1}{\rho}$ divergent? Or does it converge to any special value? Could you provide a proof to this? Here, $\rho$ is the imaginary part of the zeros of the Riemann zeta function.

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I also want to know if there is a way to approximate how many $\rho$'s there are less or equal to x (under the condition that the RH is true).

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  • $\begingroup$ Presumably you want to know about the sum where $\rho$ varies over the imaginary parts of the zeros of $\zeta$ with positive imaginary part. $\endgroup$ – Travis Willse Aug 25 '19 at 15:30
  • $\begingroup$ It is a classical result that for functions of order 1 and maximal type $\sum {\frac{1}{|\rho|}}$ is infinity, hence this sum is infinite for the RZ and non-trivial roots (by using the standard completion to an entire function etc); On the other hand, since RZ is symmetric, the sum $\sum {\frac{1}{\rho}}$ is clearly conditionally convergent if you sum it symmetrically (associating $\rho$ with say $\bar \rho$ or if you want $\gamma$ with $-\gamma$ for the imaginary parts and same real part) since the real parts are bounded so they do not really count $\endgroup$ – Conrad Aug 25 '19 at 15:31
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    $\begingroup$ Note that Riemann-von Mangoldt formula does not depend on RH. $\endgroup$ – Paolo Leonetti Aug 25 '19 at 15:43
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By the Riemann-von Mangoldt formula, we know that $$ \#\{\rho: 0< \rho \le T\} \sim c\, T\log T, $$ as $T \to \infty$, for some $c>0$.

It follows by summation formula that \begin{align} \sum_{0<\rho\le T}\frac{1}{\rho}&\asymp \frac{1}{T}\cdot T\log T-\sum_{t\le T}t\log t\left(\frac{1}{t+1}-\frac{1}{t}\right)\\ &\asymp \log T+\sum_{t\le T}\frac{\log t}{t}\\ &\asymp \log T+\log^2 T \to \infty \end{align}

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(Or, if you remember that $\sum_{n\le x}1/n\sim \log x$, then it is sufficient to note that, here, you have "more elements".)

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The Riemann-von Mangoldt function gives that the number $N(T)$ of zeros $z$ of the zeta function with $0 < \operatorname{Im} z < T$ is asymptotically distributed like $$N(T) = \frac{T}{2 \pi} \log \frac{T}{2 \pi} - \frac{T}{2 \pi} + \frac{7}{8} + \frac{1}{\pi} \arg \zeta\left(\frac{1}{2} + i T\right) + O\left(\frac{1}{T}\right),$$ and the penultimate term is $O(\log T)$.

As $T \to \infty$, the term $\frac{T}{2 \pi} \log \frac{T}{2 \pi}$ dominates, which implies the sum of reciprocals of the imaginary parts of the zeros with $0 < \operatorname{Im} z < T$ is $$\geq \frac{1}{T} \left(\frac{T}{2 \pi} \log \frac{T}{2 \pi} + O(T)\right) = \frac{1}{2 \pi} \log T + O(1) ,$$ and so diverges. With a little more work one could probably find a much better asymptotic expression for that sum of reciprocals.

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  • $\begingroup$ $N(T) = c T\log T(1+o(1)),\gamma_k = \frac1c\frac{ k}{ \log k} (1+o(1)), \sum_{k \le N(T)} \frac1{\gamma_k} = \sum_{k \le c T \log T (1+o(1))} \frac1c\frac{ \log k}{k} (1+o(1))$ $ = (1+o(1)) \sum_{k \le c T \log T } \frac1c\frac{ \log k}{k} = (1+o(1))\frac1{2c} \log^2 T$ $\endgroup$ – reuns Aug 26 '19 at 0:18
  • $\begingroup$ Thanks! And $\frac{1}{2c} = \pi$. $\endgroup$ – Travis Willse Aug 26 '19 at 0:26

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