Prove that if n is an even perfect number then $\sigma(\sigma(n)) < 6n$

Question

$\sigma(n)$ refers to the sum of all divisors function. If n is an even perfect number, then $\sigma(n) = 2n$, but why is $\sigma(\sigma(n)) < 6n$?

Answer 1

We are looking for $\sigma(2n)$, where $n$ is a perfect number. We know that all even perfect numbers are of the form $n=2^{k-1}(2^k-1)$ with $k$ prime, so we want $\sigma(2^k(2^k-1))$ We know that $\sigma$ is multiplicative, so $\sigma(2^k(2^k-1))=\sigma(2^k)\sigma(2^k-1)=(2^{k+1}-1)2^k=2n\frac {2^{k+1}-1}{2^k-1}\le 2n\frac 73\lt6n$

Answer 2

Suppose $m|2n$. Then either $m|n$ or $\frac{1}{2}m|n$. Hence the set of divisors is $\{d,2d:d|n\}$ which has sum at most $3\sum d = 6n$. Use the even property to get the strict inequality.

Edit: In case this wasn't clear, you know that at least one divisor appears doubly in the list, namely $n=2(n/2)$.

Re-edit: To make the argument more succinctly, as Erick Wong observes, note that $\sigma(mn) < \sigma(m) \sigma(n)$ if $(m,n) > 1$ (sub-multiplicativity) and $\sigma(2) = 3$.