Demonstrating that $A(N)$, the average number of positive divisors of numbers up to $N$, is approximately $\log(N)$ with error bounded by 1

Question

I would like to preface this question by saying I would prefer not to use any results from the prime number theorem (or any prime factorization for that matter) for sake of understanding.

So far, I am thinking of using mainly Riemann Sums. It is clear to me that $A(N) = \sum\limits_{i = 1}^{N} \sigma_0(i)$, however, I am having a bit more trouble trying to transform this sum into a Riemann Sum and then approximate it to be $\log(N)$ (perhaps need a refresher on Riemann Sums altogether).

Will someone please get me started?

Answer 1

For an event $X$, by $[X]$ I will mean $1$ if $X$ is true, and $0$ if it isn't. Then

$$ A(N) = \sum_{i=1}^n \sigma_0(i) = \sum_{i=1}^n |\{j : j \mid i\}| = \sum_{i=1}^n \sum_{j=1}^n [j \mid i] = \sum_{j=1}^n \sum_{i=1}^n [j \mid i] = \sum_{j=1}^n \left \lfloor \frac{n}{j} \right \rfloor, $$

and this can be approximated by a Riemann sum.