Can we find a tight approximation or a good lower bound for the following sum: $$\sum_{\text{prime } p \le n} \log_p(n)^{1/2} \ ?$$
So far, I can only come up with a very untight bound of $\ge \pi(n).$ I think the sum is at least $C \pi(n)$ for a constant $C$. I am hoping that this sum is at least $f(n) \pi (n)$ for $f$ that goes to $\infty$ with $n$. Any help is appreciated.
Abel's sum formula yields
\begin{align} \sum_{p \leqslant n} \frac{1}{\sqrt{\log p}} &= \frac{\pi(n)}{\sqrt{\log n}} + \frac{1}{2} \int_2^n \frac{\pi(t)}{t(\log t)^{3/2}}\,dt \\ &= \frac{\pi(n)}{\sqrt{\log n}} + \frac{1}{2}\int_2^n \frac{1}{(\log t)^{5/2}} + O\biggl(\frac{1}{(\log t)^{7/2}}\biggr)\,dt \\ &= \frac{\pi(n)}{\sqrt{\log n}} + \frac{n}{2(\log n)^{5/2}} + O\biggl(\frac{n}{(\log n)^{7/2}}\biggr)\,. \end{align}
Thus
$$\sum_{p \leqslant n} \sqrt{\log_p n} = \pi(n) + \frac{n}{2(\log n)^2} + O\biggl(\frac{n}{(\log n)^3}\biggr)\,.$$
So $\pi(n)$ is not a very crude estimate, it is asymptotically equal to the sum. It underestimates the sum, and we can say by how much (within the bounds we know for $\pi(x)$), but the main term is $\pi(n)$.