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given the two series

a) $ 2\sum_{\rho} \frac{1}{\rho}=A $ the sum is taken over all the zeros of the zeta function on the critical strip

b) $ \sum_{\gamma} \frac{1}{1/4+\gamma ^{2}}=S $ here the sum is taken over the imaginary part of the zeros

then is true that $ S=A $ ? I know how to calculate Z using the Hadamard product but for the second series I have no much idea

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    @Jose: Oh, you're summing the imaginary parts over both positive and negatives. I get it.2011-10-12

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This is implied by the Riemann hypothesis: Because the zeroes lie symmetrically about the real axis, the $A$ sum is really summing only the real parts of $\frac{1}{\rho}$, and $\Re(1/\rho) = \Re\left(\frac{\overline\rho}{|\rho|^2}\right) = \frac{\Re(\rho)}{\Re(\rho)^2+\gamma^2}$ So if $\Re(\rho)$ is always $1/2$, then the terms of $A$ and $S$ correspond one-to-one.

On the other hand, if the Riemann hypothesis fails, then I suppose it is still possible that the real parts magically match up such that the sums are equal even though the individual terms are not (assuming, as J.M. pointed out, that they converge at all).

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    It seems they do converge; see anon's comment in the question...2011-10-13