Does $\sum\limits_{n=1}^\infty \ln\left(\frac{p_n}{p_n - 1}\right)$ converge? Suppose $p_n$ is the $n$-th prime number. Does $\sum\limits_{n=1}^\infty \ln\left(\frac{p_n}{p_n - 1}\right)$ converge?
Where did this question arise from:
I was trying to find $\inf_{n \in \mathbb{N}} \frac{\phi (n)}{n}$, where $\phi$ is Euler totient function. If $p$ is prime, then $\frac{\phi (p^n)}{p^n} = \frac{p - 1}{p}$. As $\phi$ is multiplicative, $\phi(n) = \prod\limits_{p\mid n; p \text{ is prime}} \frac{p - 1}{p}$. That means, that $ \inf_{n \in \mathbb{N}} \frac{\phi (n)}{n} = \prod\limits_{n = 1}^\infty \frac{p - 1}{p}$. And that results in $\inf_{n \in \mathbb{N}} \frac{\phi (n)}{n} = 0$ iff $\sum\limits_{n=1}^\infty \ln\left(\frac{p_n}{p_n - 1}\right)$ diverges.

Note $\ln(p/(p-1)) = \ln(1/(1 - 1/p))$, so $\sum_{p\leq x} \ln(p/(p-1)) = \ln(\prod_{p\leq x} 1/(1-1/p))$. Intuitively, $\prod_p 1/(1-1/p) = \zeta(1) = \infty$, and this calculation can be justified. Thus, letting $x\rightarrow \infty$, we get $\sum_p \ln(p/(p-1)) = \ln(\prod_p 1/(1-1/p)) = \ln(\infty) = \infty$.
