Möbius function 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 
 A: 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}}
