It is well-known that $\sum_{n\le x}\frac{\mu (n)}{n}=o(1)$ is equivalent to the prime number theorem. But is there a direct proof for the weaker statement
$$\sum_{n\le x}\frac{\mu (n)}{n}=O(1)?$$
Namely the sum converges? Unfortunately the Dirichlet test doesn't apply here.
Clearly $\left|\sum_{n \leq x} \frac{\mu(n)}{n}\right| \leq 1$ for $x < 2$. For $x \geq 2$, we have that $$\sum_{n \leq x} \frac{\mu(n)}{n} = \frac{1}{x} \sum_{n \leq x} \mu(n) \frac{x}{n} = \frac{1}{x} \sum_{n \leq x} \mu(n) \left\lfloor \frac{x}{n}\right\rfloor + \frac{1}{x} \sum_{n \leq x} \mu(n) \left\{\frac{x}{n}\right\}.$$ Here $\lfloor x\rfloor$ denotes the integer for which $x - 1 < \lfloor x \rfloor \leq x$ and $\{x\} = x - \lfloor x\rfloor$.
For the first term, we have that $$\sum_{n \leq x} \mu(n) \left\lfloor \frac{x}{n}\right\rfloor = \sum_{n \leq x} \mu(n) \sum_{m \leq \frac{x}{n}} 1 = \sum_{mn \leq x} \mu(n) = \sum_{a \leq x} \sum_{mn = a} \mu(n).$$ Since $\sum_{mn = a} \mu(n)$ is $1$ if $a = 1$ and $0$ otherwise, we deduce that the first term is equal to $\frac{1}{x}$.
For the second term, since $\mu(n) \in \{-1,0,1\}$ for all $n \in \mathbb{N}$, we have that $$\left|\frac{1}{x} \sum_{n \leq x} \mu(n) \left\{\frac{x}{n}\right\}\right| \leq \frac{1}{x} \sum_{n \leq x} |\mu(n)| \left\{\frac{x}{n}\right\} \leq \frac{1}{x} \sum_{n \leq x} \left\{\frac{x}{n}\right\}.$$ Since $\{x\} < x$ for all $x \in \mathbb{R}$ and $\{x\} = 0$ if $x$ is an integer, we have that $$\sum_{n \leq x} \left\{\frac{x}{n}\right\} = \{x\} + \sum_{2 \leq n < x} \left\{\frac{x}{n}\right\} \leq \{x\} + \sum_{2 \leq n < x} 1 = \{x\} + \lfloor x\rfloor - 1 = x - 1.$$ We deduce that the second term is bounded by $\frac{x - 1}{x} = 1 - \frac{1}{x}$.
We conclude that $$\left|\sum_{n \leq x} \frac{\mu(n)}{n}\right| \leq \left| \frac{1}{x} \sum_{n \leq x} \mu(n) \left\lfloor \frac{x}{n}\right\rfloor\right| + \left| \frac{1}{x} \sum_{n \leq x} \mu(n) \left\{\frac{x}{n}\right\}\right| \leq \frac{1}{x} + 1 - \frac{1}{x} = 1.$$