Gold Divisibility Sum of Divisors

Hi ! For the question Sum of Divisors (, I have looked at the editorial for a hint. However, I was unable to understand the connection between the question asking for Σσ(i) and the first line of the editorial claiming that the question is equivalent to computing the the summation \sum_{i=1}^n\left\lfloor \frac{n}{i}\right\rfloor\cdot i in faster than \Theta(n) time. Could anyone explain the connection.

Let f(i)=\sum_{i=1}^n\lfloor \frac{n}{i}\rfloor \cdot i. All you need to do is verify that f(i)-f(i-1)=\sigma(i).