I am in the process of proving the prime number theorem, and I was wondering if there was any easy way of proving that $\sum_{\rho\in R} \rho^{-2}$ converges, where $R$ is the set of all non-trivial zeroes of Riemann zeta function. By easy I mean any proof that does not use the idea of the order of growth. Any simple algebraic/arithmetic proof will be most appreciated.


  • $\begingroup$ proving, what exactly? $\endgroup$ – Alex R. Apr 1 '15 at 21:06
  • $\begingroup$ Sorry, I meant the sum converges! $\endgroup$ – insignia Apr 1 '15 at 21:08

Consider $$\xi\left(s\right)=s\left(s-1\right)\pi^{s/2}\Gamma\left(s/2\right)\zeta\left(s\right)$$ that can be rewrite, using Hadamard product $$\xi\left(s\right)=\frac{e^{As}}{2}\prod_{\rho}\left(1-\frac{s}{\rho}\right)e^{s/\rho}$$ with $A=\log\left(2\pi\right)-1-2\gamma$. Now if we take logs and derive twice, we get $$\frac{d^{2}}{ds^{2}}\left(\log\left(\xi\right)\right)\left(s\right)=\sum_{\rho}\frac{1}{\left(\rho-s\right)^{2}}$$ so take $s=0$ and you get your sum. Note that with this method you can compute every $\sum_{\rho}\rho^{-n}$.


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