# Are these two series equal?

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
i know how to calculate Z... Congratulations! Thus you might wish to define Z. – Did Oct 12 '11 at 19:00
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
@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
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

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.