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

$$ \sum_{p<n}_{p\ \ prime} \frac{1}{\log{p}} $$

Additional information: Since

$$ \sum_{p<n}_{p\ \ prime} \frac{1}{\log{n}} \leq \sum_{p<n}_{p\ \ prime} \frac{1}{\log{p}} \leq \sum_{p<n} \frac{1}{\log{p}}. $$

We have that for some constants $c_1,c_2$: $$c_1\frac{n}{\log^2{n}} \leq \sum_{p<n}_{p\ \ prime} \frac{1}{\log{p}} \leq c_2 \frac{n}{\log{n}}. $$

Where 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)

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)