Is $\sum_{n=1}^{\infty} \frac{\mu(n)}{n} = 0$?

Question

Does $\sum_{n=1}^{\infty} \frac{\mu(n)}{n}$ converge and does it converge to $0$ ?. I know that $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$ converges on $Re(s) > 1$ to $\prod_{j=1}^{\infty}{(1-\frac{1}{p_j^{s}})}$ (here of course $p_1, p_2,...$ are the distinct primes) and taking $s \rightarrow 1^{+}$ suggests $\sum_{n=1}^{\infty} \frac{\mu(n)}{n} = 0$ but this is not rigorous and probably incorrect.

Answer 1

As @Conrad explained in a comment, it is indeed correct that $$\sum_{n=1}^\infty \frac{\mu(n)}{n} = 0,$$ and moreover, this result is equivalent to the Prime Number Theorem. This proof and its connection to the Prime Number Theorem have been discussed many times on this site. See for example Peter Humphries' response to this question.

For some references where you can see a detailed proof and learn a lot more about this, see Calum Gilhooley's response to this question.