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Old 2020-09-13, 04:04   #4
CRGreathouse
 
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Aug 2006

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Quote:
Originally Posted by jrsousa2 View Post
\pi(x) \sim -16\sum _{h=1}^{\infty}\frac{x^{2h+1}}{2h+1}\sum _{i=1}^h \log\zeta(2i)\sum _{v=i}^{h}\frac{(-1)^{h-v}(4\pi )^{2h-2v}}{\zeta(2v-2i)(2h+2-2v)!} \text{, if }x\text{ is sufficiently large.}
What are you taking the log of? If it's just the zeta, I don't get convergence. If it's the whole inner sum, what branch are you taking?
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