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Define the D-Ratio as the ratio of a natural number $n$ as: the sum of $n$'s Divisors, excluding 1 and $n$ divided by $n$ itself. [Thus the D-Ratio of $24$ is $$\frac{2 + 3 + 4 + 6 + 8 + 12}{24} = \frac{35}{24}.]$$

Define the cumulative D-Ratio Average to $n$ as the average D-Ratio of the natural numbers up to and including $n$. [Thus the D-Ratio Average to $10$ is $$\frac{0 + 0 + 0 + \frac{1}{2} + 0 + \frac{5}{6} + 0 + \frac{3}{4} + \frac{1}{3} + \frac{7}{10}}{10} = \frac{187}{600}.]$$

As $n$ increases, the D-Ratio Average to $n$ also increases (not with each step, but generally speaking), but it converges because -- as far as I can tell -- D-Ratios don't rise above $5.113$.

To what value does the D-Ratio Average converge?

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    $\begingroup$ You can make the $D$-ratio as large as you like, you just need enough prime factors. $\endgroup$ – Daniel Fischer Aug 27 '15 at 19:55
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These notes derive (equation ($4$) on p. $4$)

$$ \sum_{k\le n}\frac{\sigma(k)}k=\frac{\pi^2}6n+O(\log n)\;. $$

Thus for your ratio $D(n)$ we have

$$ \sum_{k\le n}\frac{\sigma(k)-k-1}k=\left(\frac{\pi^2}6-1\right)n+O(\log n)\;, $$

and dividing by $n$ shows that the average converges to

$$ \frac{\pi^2}6-1\approx0.645\;. $$

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