Proof of sum of positive divisors of $n$ (probably repeated question somewhere in the stack)

Question

If $\tau(n)$ is the number of positive divisors of $n$, including 1 and $n$, prove that

$$ S_\tau(x) = \sum_{n\leq x} \tau(n) = \sum_{j\leq x}\left[ \frac{x}{j} \right], $$

where $[x]$ is the largest integer not greater that $x$.

This is Proposition 1.2.2 in Jameson's The Prime Number Theorem. I don't understand his argument that fixing $j$ (a divisor of $n$) instead of summing over it, we only need to find $[x/j]$. A proof (if there is) using Group Theory will be ideal.

Answer 1

We have $$\sum_{n\leq x}\tau(n)=\sum_{n\leq x}\sum_{d|n}1$$ so if we let $n=dq$, we have to sum over all pairs $(d,q)$ satisfying $dq\leq x$: $$\sum_{n\leq x}\tau(n)=\sum_{dq\leq x}1 =\sum_{d\leq x}\sum_{q\leq x/d}1=\sum_{d\leq x}[x/d].$$

Answer 2

You can write $$S_{\tau} (x) = \sum\limits_{1 \leq n \leq x} \tau (n) = \sum\limits_{1 \leq n \leq x} \sum\limits_{j | n} 1 =\sum\limits_{\begin{array}{c} 1 \leq n \leq x \\ 1 \leq j \leq x \\ j | n\end{array}} 1 = \sum\limits_{1 \leq j \leq x} \sum\limits_{\begin{array}{c} n \in j \mathbb{Z} \\ 1 \leq n \leq x \end{array} } 1 = \sum\limits_{1 \leq j \leq x} \left[ \frac{x}{j} \right]$$ The crucial point is the exchange of the two sums. Then in the last step you use the fact that the number of multiples of $j$ which are between $1$ and $x$ is $\left[ \frac{x}{j} \right]$.