On page 36 of Tenenbaum's book: introduction of analytic and probabilistic number theory, 3rd ed, AMS. it says that $$\sum_{n\leqslant x} \mu(n) \left \lfloor \frac{x}{n} \right \rfloor = 1$$ suggests $$\lim_{x \rightarrow \infty} \sum_{n \leqslant x} \frac{ \mu(n)}{n} = 0$$ This $\mu(n)$ is Mobius function.
But I let $\left \lfloor \frac{x}{n} \right \rfloor = \frac{x}{n} + O(1)$, then I can only get $$\lim_{x \rightarrow \infty} \sum_{n \leqslant x} \frac{ \mu(n)}{n} = O(1)$$
For proving the claim in a easy way we have to use the PNT in the form $$\lim_{x\rightarrow\infty}\frac{M\left(x\right)}{x}=0\tag{1} $$ where $M\left(x\right)=\sum_{n\leq x}\mu\left(n\right) $ is the Mertens function. Since $\sum_{n\leq x}\mu\left(n\right)\left\lfloor \frac{x}{n}\right\rfloor =1 $ we have $$1=\sum_{n\leq x}\mu\left(n\right)\left\lfloor \frac{x}{n}\right\rfloor =x\sum_{n\leq x}\frac{\mu\left(n\right)}{n}-\sum_{n\leq x}\mu\left(n\right)\left\{ \frac{x}{n}\right\} $$ where $\left\{ x\right\} $ is the fractional part of $x$. So $$x\left|\sum_{n\leq x}\frac{\mu\left(n\right)}{n}\right|=\left|1-\sum_{n\leq x}\mu\left(n\right)\left\{ \frac{x}{n}\right\} \right| $$ $$\leq1+\sum_{n\leq x}\left\{ \frac{x}{n}\right\} =1+\left\{ x\right\} +\sum_{2\leq n\leq x}\left\{ \frac{x}{n}\right\} $$ $$\leq1+\left\{ x\right\} +\left\lfloor x\right\rfloor -1=x $$ hence $$\left|\sum_{n\leq x}\frac{\mu\left(n\right)}{n}\right|\leq1.\tag{2} $$ Now if we put $$A\left(x\right)=\sum_{n\leq x}\frac{\mu\left(n\right)}{n} $$ we have, using Abel's summation, that $$M\left(x\right)=\sum_{n\leq x}\frac{\mu\left(n\right)}{n}n=xA\left(x\right)-\int_{1}^{x}A\left(t\right)dt $$ and from $(2)$ we observe that if we fix $\epsilon>0 $ we have that exists some $c>0$ depending only on $\epsilon $ such that $\left|A\left(x\right)\right|<\epsilon $ if $x\geq c $. So $$\int_{1}^{x}\left|A\left(t\right)\right|dt=\int_{1}^{c}\left|A\left(t\right)\right|dt+\int_{c}^{x}\left|A\left(t\right)\right|dt $$ $$\leq c-1+\epsilon\left(x-c\right) $$ hence $$\left|A\left(x\right)\right|\leq\left|\frac{M\left(x\right)}{x}\right|+\frac{1}{x}\int_{1}^{x}\left|A\left(t\right)\right|dt\leq\left|\frac{M\left(x\right)}{x}\right|+\epsilon $$ and using $(1)$ and the arbitrariness of $\epsilon $ we can conclude. NOTE: this proof is classical and it is taken from Apostol, Introduction to Analytic Number Theory.