Define $\sigma(N)=\sum\_{d|n} d$. A superabundant number is a positive integer $u$ for which $\frac{\sigma(u)}{u} > \frac{\sigma(v)}{v}$ for every positive integer $v<u$.


Similarly, do there exist infinitely many positive integers $n$ for which $\frac{\sigma(n^2)}{n^2}>\frac{\sigma(m^2)}{m^2}$ for every positive integer $m<n$ ?



On superabundant-like numbers
The answer is yes because $\sigma(n^2)/n^2$ is unbounded. To see this, take the product of the first $k$ primes $n=p\_1p\_2\cdots p\_k$. We have
$$\frac{\sigma(n^2)}{n^2}=\prod\_{i=1}^k \frac{p\_i^{2}+p\_i+1}{p\_i^2}>\prod\_{i=1}^k\left(1+\frac{1}{p\_1}\right)>\sum\_{i=1}^k\frac{1}{p\_i}$$
and the final sum diverges as $k\to \infty$.


