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.
 A: Notice that
$$\ln\left(\frac{p_n}{p_n - 1}\right) = \ln p_n - \ln (p_n - 1) \approx \frac 1 {p_n}$$
by the mean value theorem, using that $\frac{d}{dx} \ln x = \frac 1 x$.
It is well known that $\sum p_n^{-1}$ diverges.
A: Observe that $$\ln\left(\frac{p_n}{p_n-1}\right)=\ln\left(1+\frac{1}{p_n-1}\right)=\frac{1}{p_n-1}+O\left(\frac{1}{(p_n-1)^2}\right).$$Thus,$$\left|\sum_{p}\ln\left(\frac{p}{p-1}\right)-\sum_p\frac{1}{p-1}\right|\ll1.$$From this, we can deduce that $\sum_p\ln\left(\frac{p}{p-1}\right)$ diverges since $\sum_p\frac{1}{p-1}$ diverges.
A: 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 $\sum_p \ln(p/(p-1)) = \ln(\prod_p 1/(1-1/p)) = \ln(\infty) = \infty$.
A: 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.
