Prove that as $x\to\infty $, $\sum\limits_{p \leq x} \frac{1}{p \log \log p} \approx \log \log \log x$

Question

Prove that as $x\to\infty$, $$\sum_{p \leq x} \frac{1}{p \log \log p} \approx \log \log \log x$$

Here sum is taken over primes.I tried to use the partial summation formula but could not choose a suitable choice of functions. It would be great if anyone can give hints. Thanks.

Answer 1

Note that holds $\underset{p\leq x}{\sum}\frac{1}{p}\sim\log\log x$ so by partial summation$$\underset{p\leq x}{\sum}\frac{1}{p\log\log p}\sim1+\int_{3}^{x}\frac{1}{\left(\log\log t\right)t\log t}dt=1-\log\log\log3+\log\log\log x\sim\log\log\log x$$ as $x\rightarrow\infty.$

Answer 2

We have $p\ln\ln p = e(n)n\ln n\cdot\ln\ln(e(n)n\ln n) $ in which by the PNT $e(n)$ goes to 1 as $n$ gets very large.

In this case $\sum\frac{1}{p\ln\ln p}\sim \int \frac{dx}{p\ln\ln p}$ and

So $$I(n) = \int_3^n \frac{dx}{p\ln\ln p} = \int \frac{dx}{e(x)x\ln x\cdot\ln\ln(e(x)x\ln x)}$$

$$\sim \int \frac{dx}{x\ln x\cdot(\ln\ln x+\ln\ln\ln x)} $$

This denomintor of the integrand is dominated by the $\ln\ln x$ term. That is,

$$x\ln x\ln\ln x+x\ln x \ln\ln\ln x \sim x\ln x\ln\ln x$$ and so

$$I(n) \sim \int \frac{dx}{x\ln x\cdot \ln\ln x}.$$

Note that

$$\frac{d}{dx}(\ln\ln\ln x )=\frac{1}{x\ln x\ln\ln x}$$ and so the integral is

$$I(n) \sim \ln\ln\ln n.$$