How can I get that $\sum_p \frac{1}{p}$ is divergent? I read this at one book. Actually I know how to show this fact, but there was a conclusion from Euler equality.

Conclusion : using $\zeta(s) = \prod(1-\frac{1}{p^{s}})^{-1}$ we could found out this : $\sum \frac{1}{p} \rightarrow \infty$. 

I thought about Cauchy inequality, but I got only upper bound : $\sum (1-\frac{1}{p}) \ge \prod(1-\frac{1}{p})$. 
 A: From Euler's product we have that for any $s>1$
$$\log\zeta(s) = \sum_{p}-\log\left(1-\frac{1}{p^s}\right)=\sum_p\left(\frac{1}{p^s}+\frac{1}{2p^{2s}}+\frac{1}{3p^{3s}}+\ldots\right) $$
holds, but over the interval $(0,1/2]$ the function $f(x)=-\log(1-x)-x$ is bounded in absolute value by $x^2$, hence
$$ \log\zeta(s) = \sum_{p}\frac{1}{p^s}+O(1). $$
By exploiting Abel's summation formula it is not difficult to prove that $\zeta(s)$ behaves like $\frac{1}{s-1}$ in a right neighbourhood of $s=1$, hence $\lim_{s\to 1^+}\log\zeta(s)=+\infty$ and by the previous formula $\sum_{p}\frac{1}{p}$ cannot be convergent.
A: By using prime number theorem we know,
Let $\pi(x)$ be the prime-counting function that gives the number of primes less than or equal to $x$, for any real number $x$. Then 
$$\lim_{x\to \infty}\frac{\pi(x)}{\frac{x}{\ln x}}=1$$
Hence for the $n$ th prime number $p_n$, we have 
$$\lim_{n\to\infty}\frac{p_n}{n\log n}=1$$
But we know when $\lim_{n\to\infty}\frac{a_n}{b_n}=1$ then $\sum a_n$ and $\sum b_n$ have same behaviour , hence since $\sum\frac{1}{n\log n}$ diverges by using Cauchy test, so $\sum\frac{1}{p_n}$ diverges
I know it from the book of T.M.Apostol since my BSc 
