Convergence of a Formal Series Involving the Mobius Function

Question

Let $\mu: \mathbb{N} \to \{0, \pm 1\}$ be the Mobius function. It is not hard to show that, formally, $$ \prod_{n \geq 1} (1 - x^n)^{-\mu(n) / n} = e^x $$ by taking logarithms and expanding using the power series of log. My question is, when does the left hand side actually converge to $e^x$?

Answer 1

Note first that if $|x|>1$ then you run into branch-cut issues for even $n$. In the case of $|x| = 1$, the left-hand side is zero and the right-hand-side is not, so there's no convergence there.

We thus can restrict to the case of $|x| < 1$. Taking $\log$, we're looking to see if the series $\sum_{n \geq 1} -\frac{\mu(n)}{n}\log(1 - x^n)$ converges. In fact, it's not too hard to show that this converges absolutely for $|x| < 1$; to do this, note that $|\log(1 - y)| < 2 |y|$ for $|y|$ sufficiently small. This then allows us to eventually bound the series above by a geometric series. Since it converges absolutely, we're free to make the manipulations to show that $$\sum_{n\geq 1} -\frac{\mu(n)}{n}\log(1 - x^n) = x\,.$$