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Section 19.3 The Size of the Sum of Divisors Function

For the rest of this chapter, we will focus on \(\sigma_1=\sigma\) itself, since the sum of divisors function has a deep richness of its own. We could ask questions about evenness, other patterns, and so forth.
This short section asks a particularly interesting question. Try the following interactive cell.
This table helps you see possibilities for the relative size of \(\sigma(n)\) with respect to \(n\) itself. Alternately, we have the following.

Question 19.3.1.

For any given \(n\text{,}\) what is the constant \(C_n\) such that \(\sigma(n)=C_n\cdot n\text{?}\) How big can this get?
The spread of these ratios, for \(n\) under one hundred fifty, certainly goes both above and below \(2\text{.}\) If you look carefully, you will see that only one of the numbers above has a sum of divisors without 1 or 2 as the integer part. What is it?
Instead of simply trying larger and larger input numbers, we might use a little theory to get a higher ratio. To wit, if a number has lots of small prime divisors, we might think it has lots of factors. So taking big powers of these would have even more small prime divisors and might get us big ratios.
You’ll notice that although we quickly get a ratio above 3 (so that \(\sigma(n)\gt \)), we don’t seem to get much further. Why?
A helpful thing to think about with this is the following rewrite, using the formula for \(\sigma(n)\) with the usual writing of \(n=\prod_{i=1}^k p_i^{e_i}\text{:}\)
\begin{equation*} \frac{\sigma(n)}{n}=\frac{\prod_{i=1}^k \left(\frac{p_i^{e_i+1}-1}{p_i-1}\right)}{\prod_{i=1}^k p_i^{e_i}} = \prod_{i=1}^k\frac{p_i-\left(1/p_i^{e_i}\right)}{p_i-1} \approx \prod_{i=1}^k\frac{p_i}{p_i-1} \end{equation*}
Based on this, we should expect this approximation to be very close when \(e_i\) are all quite large. Then for large numbers, since \(\frac{p}{p-1}>1\text{,}\) if we multiply by enough of these we will get very large numbers and so \(\sigma(n)/n\) will be greater than any given \(C\text{,}\) and then \(\sigma(n)>Cn\text{.}\)
Of course, \(p=2\) is the best for this since \(\frac{2}{2-1}=2\text{,}\) but the other primes will hopefully be useful for this as well. For instance, \(n=2^{10} 3^{10}\) will have
\begin{equation*} \sigma(n)/n=\frac{2-1/2^{10}}{2-1}\cdot \frac{3-1/3^{10}}{3-1} \approx \frac{2}{2-1}\cdot \frac{3}{3-1}=3 \end{equation*}
so certainly \(\sigma(6^{10})\) will be nearly \(3\cdot 6^{10}\text{.}\)
If we multiply it by 5 as well that should do it, and that gives the results we saw in the previous cell:
\begin{equation*} \frac{2-1/2^{10}}{2-1}\cdot \frac{3-1/3^{10}}{3-1}\cdot \frac{5-1/5}{5-1}\approx\frac{2}{2-1}\cdot \frac{3}{3-1}\cdot \frac{5}{5-1}=2\cdot \frac{3}{2}\cdot \frac{5}{4}=\frac{15}{4}=3.75 \end{equation*}
We can check out some of these ideas, and how much bigger we can get.
Continuing this for more primes suggests the following.
The argument outlined above is not completely rigorous, but is good enough for now. Trying to prove it this way could bring the distribution of primes to the table, so doing so might not be trivial. (As it happens, one can prove this in a very elementary way; see [E.4.5, Section 3.6].)