**Question:** What is the order of magnitude of the following sum?


$$ \sum\_{\substack{p<n\\\text{$p$ prime}}} \frac{1}{\log{p}} $$ 


**Additional information**: Since 


$$ \sum\_{\substack{p<n\\\text{$p$ prime}}} \frac{1}{\log{n}} \leq \sum\_{\substack{p<n\\\text{$p$ prime}}} \frac{1}{\log{p}} \leq \sum\_{p<n} \frac{1}{\log{p}}, $$


we have that, for some constants $c\_1,c\_2$,  

$$c\_1\frac{n}{\log^2{n}} \leq \sum\_{\substack{p<n\\\text{$p$ prime}}} \frac{1}{\log{p}} \leq c\_2 \frac{n}{\log{n}}. $$


Here, the asymptotics on the left hand side came from the prime number theorem, and on the right hand side from the asymptotic expansion of the logarithmic integral function. 


(See: <http://en.wikipedia.org/wiki/Logarithmic_integral_function#Asymptotic_expansion>) 



Order of magnitude of $\sum \frac{1}{\log{p}}$
The contribution of the primes $p\leq n/\log ^3 n$ is clearly $O(n/\log^3 n)$. For the remaining primes we have
$$ \log n-3\log\log n <\log p<\log n,$$
$$ \frac{1}{\log p}=\frac{1}{\log n}+O\left(\frac{\log\log n}{\log^2 n}\right), $$
so by the Prime Number Theorem their contribution is
$$ \left(\frac{n}{\log n}+O\left(\frac{n}{\log^2 n}\right)\right)\left(\frac{1}{\log n}+O\left(\frac{\log\log n}{\log^2 n}\right)\right)=\frac{n}{\log^2 n}+O\left(\frac{n\log\log n}{\log^3 n}\right).$$
Altogether we see that
$$ \sum\_{p<n} \frac{1}{\log{p}} =\frac{n}{\log^2 n}+O\left(\frac{n\log\log n}{\log^3 n}\right).$$


Better bounds can be obtained by partial summation, namely
$$ \sum\_{p\leq n} \frac{1}{\log{p}} = \frac{\pi(n)}{\log n}+\int\_2^n\frac{\pi(x)}{x\log^2 x}dx. $$ 


