For any natural number $x$, determine the sum; $$ \sum_{\substack{ n\leq x }} \mu(n)\left\lfloor \frac{x}{n} \right\rfloor.$$ (Hint: Use $\lfloor x \rfloor=\sum_{\substack{ k\leq x }}1.$)
Here is my start: $$=\sum_{\substack{ n\leq x }} \mu(n)\sum_{\substack{ k\leq \frac{x}{n} }}1$$ if we change the order of summations then $$=\sum_{\substack{ k\leq x }}1 \sum_{\substack{ n\leq \frac{x}{k} }}\mu(n)$$ But I couldn't go further
Using the given hint, $\let\leq\leqslant$ $$\begin{aligned} \sum_{n\leq x}\mu(n)\left\lfloor\frac xn\right\rfloor =&\sum_{n\leq x}\mu(n)\sum_{\substack{k\\ k\leq\frac xn}}1\\ =&\sum_{n\leq x}\mu(n)\sum_{\substack{k\\ kn\leq x}}1\\ \overset{(kn=m)}=&\sum_{n\leq x}\mu(n)\sum_{\substack{m\\ m\leq x\\n\mid m}}1. \end{aligned}$$ Note the changes under the right summation sign. In the last step $kn$ was replaced by $m$, giving $m$ the extra condition that it should be divisible by $n$.
Now we change the order of summation: $$\begin{aligned} \sum_{n\leq x}\mu(n)\sum_{\substack{m\leq x\\n\mid m}}1 &\overset{(n\;\leftrightarrow\;m)}=\sum_{m\leq x}1\sum_{\substack{n\\n\leq x\\n\mid m}}\mu(n)\\ &=\sum_{m\leq x}1\sum_{n\mid m}\mu(n). \end{aligned}$$ Note that in the last step we omitted the condition $n\leq x$ because it was superfluous.
Because $\sum_{n\mid m}\mu(n)=0$ if $m>1$, this simplifies to $1$.
As you can see, using the hint in the form it was given is a matter of manipulating conditions under the summation sign. In this kind of sums I recommend you to write the conditions as explicit as possible. That's what I do too, and it helps to avoid making mistakes when changing the order of summation. To enable this is $\LaTeX$, you can use \sum_{\substack{info \\ more info \\ some more info}}
