# Is the sum of the reciprocals of imaginary parts of the zeros of the zeta function divergent?

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.

Edit 1:

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).

• Presumably you want to know about the sum where $\rho$ varies over the imaginary parts of the zeros of $\zeta$ with positive imaginary part. – Travis Willse Aug 25 '19 at 15:30
• 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 – Conrad Aug 25 '19 at 15:31
• Note that Riemann-von Mangoldt formula does not depend on RH. – Paolo Leonetti Aug 25 '19 at 15:43

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".)

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.

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