Interesting convergence in divisor sums up to $10^k$
This is all due to the fact that $$ \frac{1}{x}\sum_{n\le x} \sigma(n) = \frac{\pi^2}{12}x + O(\log x). $$ See these notes by Carl Pomerance (equation (5) if you're in a hurry).
This is all due to the fact that $$ \frac{1}{x}\sum_{n\le x} \sigma(n) = \frac{\pi^2}{12}x + O(\log x). $$ See these notes by Carl Pomerance (equation (5) if you're in a hurry).