So far i've tried looking at an equation which is true for perfect numbers:
$\sigma(n) = 2n$. where $\sigma(n) = \sum_{d|n} d$
i started looking at a odd perfect number which has 1 prime divisor say $p$, then:
$\sigma(n) = \sum_{i \in \mathbb{N}} p_i$, where all $p_i$'s are the same except for $p_1 = 1$. But then $\frac{\sigma(n)}{n} = 2$ right...?
Any helpfull tips here?
Kees
Hints:
(1) $\sigma(p^a)=1+p+p^2+...+p^a=\frac{p^{a+1}-1}{p-1}$. And so $\frac{\sigma(p^a)}{p^a}<\frac{p}{p-1}$.
(2) So $\frac{\sigma(p^a\cdot q^b)}{p^a\cdot q^b}<\frac{p}{p-1}\cdot\frac{q}{q-1}$
(3) $\frac{p}{p-1}$ is decreasing (i.e., it is larger for small primes).