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.

Another way, it is known that
$$\ln{(1+x)}\geq \frac{x}{1+x}, \forall x>-1$$
and 
$$\ln{\left(\frac{p_n}{p_n-1}\right)}=\ln{\left(1+\frac{1}{p_n-1}\right)}\geq \frac{\frac{1}{p_n-1}}{1+\frac{1}{p_n-1}}=\frac{1}{p_n}$$
thus
$$\sum\limits_{n}\ln{\left(\frac{p_n}{p_n-1}\right)} \geq \sum\limits_{n}\frac{1}{p_n}$$
which diverges.
