Sum of reciprocals of primes: easy proof that $\sum_{p\leq x} \frac{1}{p} < C\log\log x$ for some constant $C$?

Question

$\sum_{p\leq x} \frac{1}{p} \sim \log\log x$ follows from the PNT, and there's also a short proof not relying on the PNT. However that proof still requires some nontrivial steps. I was wondering if there is a simpler proof which shows the weaker $$C_1\log\log x < \sum_{p\leq x} \frac{1}{p} < C_2\log\log x$$ for some constants $C_1, C_2$.

What made me wonder is that if I start from a proof that $\sum_p \frac{1}{p}$ diverges (which I take from Bateman & Diamond's ANT book): $$\sum_{n\leq x} \frac{1}{n} < \prod_{p\leq x} \left(1-\frac{1}{p}\right)^{-1}<\prod_{p\leq x} e^{2/p} = \exp(2\sum_{p\leq x}\frac{1}{p})$$ and take the $\log$ of each side I get $$\sum_{p\leq x} \frac{1}{p} > \frac{1}{2}\log\left(\sum_{n\leq x}\frac{1}{n}\right) > \frac{1}{2}\log\log x$$ which gives one side of the required inequality. Is there a way to get a similarly easy proof of the other side?

Relatedly, is there a proof of $\sum_{p\leq x}\frac{1}{p} \sim \log\log x$ which is simpler than the one in Merten's theorems (and which does not rely on the PNT)? Mertens gives more, $$\sum_{p\leq x}\frac{1}{p}=\log \log x + M + O\left(\frac{1}{\log n}\right)$$ which is why I ask if there is a simpler proof which does not give the asymptotic term.

Answer 1

Okay here is a really easy way to do this. This solution DOES NOT USE ANYTHING about the distribution of primes.

Consider $A=\sum_{p<n}1/p$. Now consider $A^k=\sum_{p_1,p_2,\dots,p_k<n}\frac{1}{p_1p_2\dots p_k}$. By Fundamental theorem of Arithmetic, any integer can be uniquely factorized, thus, any $1/m$ are calculated at most $k!$ times. Furthermore, the integers $p_1\dots p_k\le n^k$. Therefore, we know that $A^k=\sum_{p_1,p_2,\dots,p_k}\frac{1}{p_1p_w\dots p_k} \le k!\sum_{i=1}^{n^k}1/i\le k!\times 2k\ln n$. Then we do $1/k$ power at both sides, we have $A\le (k!2k\ln n)^{1/k}\le 10k(\ln n)^{1/k}$. This can be proved by noticing the Stirling's Formula. Now we take $k$ to be $\lfloor\ln \ln n+1\rfloor$, and we have $10k(\ln n)^{1/k}\le 10(\ln\ln n+1)\times e$. Now we take $n\ge 3$ so we can take $C=10e\times(1+\ln\ln 3)/(\ln\ln3)$

Answer 2

Chebyshev was able to prove by elementary means (although it's still somewhat tricky) that there exist positive constants $c_1, c_2$ such that $$c_1\frac{x}{\log x} \lt \pi(x) \lt c_2\frac{x}{\log x}$$ for sufficiently large $x$.

Using the Abel summation formula and the previous result we get:

$$\sum\limits_{p \leq x}\frac{1}{p} = \sum\limits_{n \leq x}\frac{\mathbb{1}_P(n)}{n} = \sum\limits_{n \leq x}\frac{\pi(n)-\pi(n-1)}{n} = \frac{\pi(x)}{x} + \int\limits_1^x\frac{\pi(u)}{u^2}du$$

$$ \leq \frac{c_2}{\log x} + c_2\int\limits_1^x\frac{1}{u\log u}du$$

$$ \leq \frac{c_2}{\log x} + c_2\log\log x$$

$$ \leq C_2\log\log x$$

for some other constant $C_2$ and sufficiently large $x$.