Prime reciprocals series proof

Question

I am currently struggling with the proof of the prime reciprocals series divergence. I've already proved that :

$$\prod_{k=1}^n \frac{1}{1-1/p_k} \longrightarrow +\infty$$

Let $n$ be an integer. How to find $K$ such that :

$$\sum_{k=1}^K \frac{1}{p_k} \geq \sum_{k=1}^n \ln\bigg{(}\frac{1}{1-1/p_k}\bigg{)}$$

It's the last step to demonstrate the divergence of this series, but I can't manage to do so. Could you help me ?

Answer 1

Since $\log (1 + x) \le x$ for all $x>0$ and $\frac{1}{{1 - \frac{1}{x}}} \le 1 + \frac{2}{x}$ for $x \ge 2$, one has $$ \frac{1}{2}\log \left( {\prod\limits_{k = 1}^n {\frac{1}{{1 - 1/p_k }}} } \right) = \frac{1}{2}\sum\limits_{k = 1}^n {\log \frac{1}{{1 - 1/p_k }}} \le \frac{1}{2}\sum\limits_{k = 1}^n {\log \left( {1 + \frac{2}{{p_k }}} \right)} \le \frac{1}{2}\sum\limits_{k = 1}^n {\frac{2}{{p_k }}} = \sum\limits_{k = 1}^n {\frac{1}{{p_k }}} . $$ But as you have already shown, the left-hand side diverges as $n\to +\infty$.

Answer 2

$\ln (\frac 1 {1-t})=\ln (1+\frac t {1-t}) \leq \frac t {1-t}$. From this you get divergence of $\sum_k \frac 1 {p_k-1}$. Since $\frac {p_k} {p_k -1} \to 1$ it follows that $\sum_k \frac 1{p_k}$ is also divergent.