Robert Gerbicz wrote: Suppose indirectly that there are only finite number of super-abundant numbers and let $ N$ is the largest. Let $ p>N$ prime number, then $ gcd(N,p)=1$ so $ {\sigma(N*p)}={\sigma(N)}*(p+1)$ from this \[ \frac{{\sigma(N*p)}}{N*p}=\frac{{\sigma(N)}*(p+1)}{N*p}>\frac{\sigma(N)}{N}\] So there is at least one $ m$ number greater than N for that \[ \frac{\sigma(m)}{m}>\frac{\sigma(N)}{N}\], if K is the smallest among them, then K is trivially a super-abundant number.

In fact, it is a trivial corollary to this simple (derived from the above) Theorem. Let $(a_n)_{n\geq 1}$ be a sequence of positive real numbers. Call a term $a_m$, $m\geq 2$, to be a gigant if $a_m > a_k$ for all $1\leq k<m$. Then, if there exists no largest element, there are infinitely many gigants. Proof. Assume $a_N$ is the largest gigant. Since there exists no largest element, the set $\{m \mid a_m > a_N\}$ is non-empty. Its least element $M$ then yields a gigant $a_M$, since $a_k \geq a_M$ for some $1\leq k < M$ would force $a_k > a_N$, thus $N<k<M$, contradicting the minimality of $M$. All that was left then was to show that the sequence $\left (\dfrac {\sigma(n)} {n}\right )_{n\geq 1}$ has no largest element; in fact it is even upper unbounded. Indeed, for $n= \prod_{k=1}^s p_k$ we have $\dfrac {\sigma(n)} {n} = \prod_{k=1}^s \left (1+ \frac {1} {p_k}\right ) > 1+ \sum_{k=1}^s \frac {1} {p_k} \to \infty$ when $s\to \infty$.