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It was not hard to google that the simpler sum $\displaystyle \sum_{p < N} \frac{1}{p} = c + \log \log N $ so it is divergent.

The sum of reciprocals of primes squared also converges $\displaystyle \sum_p \frac{1}{p^2} = 0.4522\dots$


Does $\displaystyle \sum_{p} \frac{1}{p} \left\{ \frac{N}{p} \right\} $ stay bounded as $N$ gets large? What about $\displaystyle \sum_{p,q } \frac{1}{pq} \left\{ \frac{N}{pq} \right\} $ for large $N$?

Maybe it's necessary to say $p,q < N$. This is related to my earlier question on the Euler $\phi$-function

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    $\begingroup$ The first sum is asymptotic to $(1-\gamma)\log \log N$, where $\gamma$ is Euler's constant, and the second sum is asymptotic to $c_2(\log \log N)^2$ for some constant $c_2$. I am very curious though, how does this relate to your earlier question? $\endgroup$ – Eric Naslund May 3 '12 at 18:21

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