Bound of the sum $\sum_{p\le n}\frac{1}{\log(p)}$ While doing a sum I came to the sum $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}$.
Where the $\log$ is the natural logarithm.
It was easy to prove that $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}<\pi(n)+\text{const.}$.
Thus $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}=o\left(\frac{n}{\log(n)}\right)$.
Using my computer I can see that $\displaystyle\sum_{p\le n}\dfrac{1}{\log(p)}=o\left(\log(n)^{3/2}\right)$.
I can go further and see that it is $o\left(\log(n)^{1.4}\right)$. It may be true that $o\left(log(n)^{1.35}\right)$.
(I have checked $n$ till $10^{19}$. My C++ code can't be trusted anymore because $64$ bit integers can't take in more values and precision of long double is up to like $16$ digits.)
My conjecture is that it is $O(\log(n))$. Is this true? Are there any other bounds? I don't want advanced proofs as I am a novice in Maths. If there is any bounds just post them here without the proofs and with some comments, if necessary.
 A: Just split the sum into $\sum_{p \le (n/\log^2 n)}$ + $\sum_{(n/\log^2 n) < p \le n}$.  The first piece is easily $O(n/\log^3 n)$ and the second piece is asymptotic to $n/\log^2 n$, so without any "advanced" proofs like partial summation one has that the sum is asymptotic to $n/\log^2 n$.
Some additional detail: note that for any $(n/\log^2 n) < p \le n$, $\log p$ is very close to $\log n$.  Specifically, we have $\log n - 2 \log \log n < \log p \le \log n$ so that $\log p = (1 - o(1)) \log n$ and $\frac1{\log p} \sim \frac1{\log n}$ (at least for $n$ large enough that the $o(1)$ is bounded away from $1$).  Meanwhile the number of terms in the sum $\sum_{(n/\log^2 n) < p \le n}$ is $\pi(n) - \pi(n/\log^2 n)$, where $\pi(n)$ is asymptotic to $n/\log n$ and $\pi(n/\log^2 n)$ may be trivially bounded by $n/\log^2 n = o(n/\log n)$.  Thus the second piece is asymptotic to $\frac1{\log n}\cdot\frac{n}{\log n}$.
A: By partial summation, we have
$$
\sum_{p\le x}{1\over\log p}={\pi(x)\over\log x}+\int_2^x{\pi(t)\over t\log^2t}\mathrm dt
$$
Due to the prime number theorem that
$$
\pi(x)={x\over\log x}+\mathcal O\left(x\over\log^2x\right)
$$
we have
$$
\sum_{p\le x}{1\over\log p}={x\over\log^2x}+\mathcal O\left(x\over\log^3x\right)+\mathcal O\left(\int_2^x{\mathrm dt\over\log^3t}\right)
$$
By partitioning the remaining integral into
$$
\int_2^{\sqrt x}+\int_{\sqrt x}^x
$$
it easily follows that
$$
\int_2^x{\mathrm dt\over\log^3t}\asymp{x\over\log^3x}
$$
Consequently, we have
$$
\sum_{p\le x}{1\over\log p}={x\over\log^2x}+\mathcal O\left(x\over\log^3x\right)
$$
