The average number of divisors of positive integers from 1 to n ...

$\left( \begin{array}{ccccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)$

where the columns of the table have ones where the divisors are.

Let us read the table row-wise instead. The total number of divisors for the positive integers from 1 to $n$ is $\sum _{i=1}^n \left\lfloor \frac{n}{i}\right\rfloor$ . The average therefore is $\frac{1}{n}\sum _{i=1}^n \left\lfloor \frac{n}{i}\right\rfloor$ . The average is less then or equal to $\frac{1}{n}\sum _{i=1}^n \frac{n}{i}$ where the error in each term being added is less than 1. This can be rewritten as $\sum _{i=1}^n \frac{1}{i}$ . These are harmonic numbers. Therefore, $H_n-1\lt \overline{t}_n \leq H_n$ . Comparing the upper and lower Riemann sums at positive integer boundaries to $\int_1^n \frac{1}{t} dt=log_{\mathbb{e}} n$ gives $\log_{\mathbb{e}} n+\frac{1}{n}< H_n<\log_{\mathbb{e}} n+1$ . In the limit as $n\to\infty$ , $\log_{\mathbb{e}} n< H_n<\log_{\mathbb{e}} n+1$ . In conclusion, $\left|\overline{t}_n-\log_{\mathbb{e}} n\right|\leq 1$ . A deviation of 1 is achieved at $n=1$ .