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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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Do those things even converge? We don't even know all the zeroes... –  J. M. Oct 12 '11 at 18:39
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i know how to calculate Z... Congratulations! Thus you might wish to define Z. –  Did Oct 12 '11 at 19:00
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you can take the logarithmic derivative inside the Hadamard product for the Riemann Xi function :) , in fact in MATHWORLD there is a description of the sum over Riemann zeros mathworld.wolfram.com/RiemannZetaFunctionZeros.html equation (4) to (10) –  Jose Garcia Oct 12 '11 at 19:30
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@J.M. Jose is right, the sum in (a) converges and can be found exactly (as in the links). That's been known since Riemann. OP: Why do you have that extra factor of $2$ in the first sum? –  anon Oct 12 '11 at 21:18
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the problem is the following 'anon' according to Mathworld the sum $\sum_{\rho} \frac{1}{\rho} $ converges to $ \frac{2+\gamma-log4\pi}{2} $here 'gamma' is the Euler mascheroni constant .. however if we assume RH the sum $\sum_{t} \frac{4}{1+4t^{2}} $ is equal to $ 2+\gamma-log4\pi $ –  Jose Garcia Oct 12 '11 at 21:37

1 Answer 1

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... –  J. M. Oct 13 '11 at 3:15

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