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I would like an asymptotic estimate for $\displaystyle\sum_{p\leq x} \frac{1}{p\log p}$ ($p$ prime).

Specifically I am trying to show that $L^2\sum_{L^2\leq p\leq\exp(\log^2 L)}\frac{1}{p\log p}$, where $L:=\sqrt{\log N\log\log N}$, is dominated by $\log N$ for large $N$.

I have tried using summation by parts but can’t get anything useful out of it. Help would be appreciated.

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  • $\begingroup$ If nobody answers this in a couple of hours, I’ll work it out and give you a bound. $\endgroup$ – Clayton Jan 4 '20 at 0:11
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Summation by parts + the PNT $$\sum_{p\le x}1\sim \sum_{n\le x} 1/\log n$$ gives $$\sum_{p\ge x} \frac1{p\log p}\sim \sum_{n\ge x} \frac{1}{n\log^2 n} \sim \frac{1}{\log x}$$ You can replace the PNT by Mertens theorem.

In particular $$A=\sum_p \frac1{p\log p}$$ converges and $$\sum_{p\le x} \frac1{p\log p}= A-\frac{1}{\log x}(1+o(1))$$ Are you asking if there is an alternative expression for $A$ ?

The next error term depends on the Riemann hypothesis.

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