Möbius function $\mu(n)$ is defined as $\mu(n) = (-1)^k$ when $n$ is square free and $n$ contains $k$ distinct prime factors, and 0 otherwise. Also $\mu(1)=1$.
There are nice properties of this function like $$\sum_{d|n} \mu(d) = 0$$ unless $n=1$.
Let us consider $$S(x) = \sum_{n\leq x} \frac{\mu(n)}{n}$$ We see that $S(1)=1$ and $S(2) = \frac{1}{2}$ and so on. What I observed by examples is $|S(x)| \leq 1$ for all $x$ and equality is satisfied only when $x=1$ but I am not able to prove or disprove this statement for any arbitrary $x$. If this is true then there is a possibility that $\sum_{n=1}^{\infty} \frac{\mu(n)}{n}$ converges.
As mentioned in one of the comments $$\sum_{n \leq x} \mu(n) \left\lfloor \frac{x}{n} \right\rfloor = 1$$ is helpful in showing your inequality is true.
Observe that
$$ 1 = \sum_{n\leq x} \mu(n) \left\lfloor \frac{x}{n} \right\rfloor = \sum_{n\leq x} \mu(n) \left( \frac{x}{n} - \left\{ {\frac{x}{n}} \right\}\right) = xS(x) - \sum_{n\leq x} \mu(n) \left\{ {\frac{x}{n}} \right\} $$
where $\left\{ y \right\}$ is fractional part of $y$ which is always between 0 and 1.
So $$x|S(x)| \leq 1 + \sum_{n \leq x} \left\{ {\frac{x}{n}} \right\} \leq 1 + \left\{ x \right\}+ \sum_{2 \leq n \leq x} \left\{ {\frac{x}{n}} \right\} \le 1 + \left\{ x \right\}+ \lfloor x\rfloor -1 = x$$
Dividing by $x$ we get your strict inequality.