4
$\begingroup$

In my Number Theory skript it says:

  1. 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 $p\le x$.

  2. By showing that there are at most $\sum_{x

  3. 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?

  • 1
    I think the second inequality in 2. is backwards. It should read "By showing that there are at most $\sum{x numbers $m\le n$ which are divisible by at least one prime with $p \ge x$."2012-03-15

1 Answers 1