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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...2011-10-12
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    *i know how to calculate Z*... Congratulations! Thus you might wish to define Z.2011-10-12
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    oh sorry J.M it was my mistake :) i wanted to say 'i know how to calculate A (the first sries....2011-10-12
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    You do? Really? Nobody's found all the zeroes... how'd you swing that?2011-10-12
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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 http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html equation (4) to (10)2011-10-12
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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?2011-10-12
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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 $2011-10-12
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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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