In my Number Theory skript it says:
By showing that there are at most $(1+\frac{\log n}{\log 2})^{\pi(x)}$ numbers with $m\le n$ which are divisible only by prime numbers $p\le x$.
By showing that there are at most $\sum_{x
numbers $m\le n$ which are divisible by at least one prime with $p\le x$.
With 1. and 2. we can conclude that $\sum_{p\in \mathbf{P}} \frac{1}{p}$ is divergent by otherwise choosing x with $\sum_{p>x} \frac{1}{p} < \epsilon \le \frac{1}{2}$ and then $n\le (1+\frac{\log n}{\log 2})^{\pi (x)} + \epsilon n$ follows for all n.
How can we show 1. and 2. ? And how does my professor conclude in 3?
numbers $m\le n$ which are divisible by at least one prime with $p \ge x$."
– 2012-03-15