I'm reading a paper which asserts the following:
$$\sum_{n/2 < p \leq n} \frac{1}{p} \sim \frac{\log 2}{\log n}$$
follows from prime number theorem, where the sum is taken over $p$ prime. What is the explanation for this? I tried working it out using
$$\pi(x) \sim \frac{x}{\log(x)}$$
but didn't get anything useful. Sorry if this question is easy, I don't have much experience with ANT.
We need a more information than just the PNT. What we need the following refined form of PNT.
$$ \pi(x) = \frac{x}{\log x}+O\left(\frac x{(\log x)^2}\right). $$
We can write your sum as (apply partial summation) $$ \int_{n/2}^n \frac{d\pi(t)}t = \frac{\pi(n)}{n} - \frac{\pi(n/2)}{n/2} + \int_{n/2}^n \frac{\pi(t)}{t^2} dt $$
Now, using the form of PNT above, we have $$ \int_{n/2}^n \frac{d\pi(t)}t = \int_{n/2}^n \frac{\pi(t)}{t^2} dt + O\left(\frac1{(\log n)^2}\right). $$
Using the form of PNT again, we have $$ \int_{n/2}^n \frac{\pi(t)}{t^2} dt=\log\log n - \log\log (n/2) +O\left(\frac 1{(\log n)^2}\right)$$
Since $\log\log n - \log\log(n/2) \sim \frac{\log 2}{\log n}$, we have the result.