Lower bound for sum of reciprocal of primes

Question

In page 70 of the book Approximate formulas for some functions of prime numbers by J. Barkley Rosser and Lowell Schoenfeld, it is given that \begin{equation} \log\log x < \sum_{p\leq x}\frac{1}{p}\quad \text{for}\, x>1 \end{equation} How can one prove it? Is there an elementary proof of this? No proof has been mentioned in the book.

Answer 1

We can argue as follows. A truncation of the Euler product gives

$$\prod_{p \le x} \left( \frac{1}{1 - \frac{1}{p}} \right) > \sum_{n=1}^x \frac{1}{n} = H_x > \log x$$

(since the LHS is exactly the sum of $\frac{1}{n}$ over all $n$ whose prime factors are $\le x$). Taking logs gives

$$\sum_{p \le x} - \log \left( 1 - \frac{1}{p} \right) > \log H_x > \log \log x$$

(all logs are natural). Taylor's theorem with remainder on the interval $[0, \frac 1 2]$ gives that if $0 \le x \le \frac 1 2$ then

$$\frac{x^2}{4} \le \log (1 - x) + x \le x^2$$

so $\log (1 - x) \le -x + x^2$ (we only need this half of the bound) on $[0, \frac 1 2]$ which gives

$$\sum_{p \le x} \left( \frac{1}{p} + \frac{1}{p^2} \right) \ge \sum_{p \le x} - \log \left( 1 - \frac{1}{p} \right) > \log \log x.$$

Since $\sum_{p \le x} \frac{1}{p^2} < \sum_p \frac{1}{p^2} < \sum_n \frac{1}{n^2} = \frac{\pi^2}{6}$ this gives

$$\sum_{p \le x} \frac{1}{p} > \log \log x - \frac{\pi^2}{6}$$

which is not quite as good as what you asked for but pretty good; this argument requires no particularly hard tools. A slightly better version of this bound is given on Wikipedia but it doesn't get the constant all the way down to zero. There are several places in this argument where the bounds can be tightened up.