How to prove divergence of $\sum_{p \text{ prime}} \frac 1 p$?

Question

I want to prove that $$\sum_{p \text{ prime}} \frac 1 p$$ diverges. I checked the wikipedia page about that: Divergence of the sum of the reciprocals of the primes, but I wanted something simple. So I just used the fact that the next prime is always greater than the last one to show that: $|\frac{a_{n+1}}{a_n}| > 1,$ where $a_n$ is $1$ over the $n$-th prime, so the sum must diverge. But I am not sure this is ok, can anybody tell me?

Answer 1

The answer is NO:

Let $a_n= \frac{1}{p_n}$, $p_n < p_{n+1}$ Then actually $|\frac{a_{n+1}}{a_n}| < 1$

Answer 2

Actually $$\left|\frac{1/p_{n+1}}{1/p_{n}}\right|<1$$

In the limit, all you get that:

$$\lim_{n\to\infty}\left|\frac{1/p_{n+1}}{1/p_{n}}\right|\le 1 $$

So even if your inequality WAS true, it wouldn't be enough to prove divergence.