for which conditions of postive integer $n, m >0$ :$\dfrac{\sigma{(n)}}{n}\leq\dfrac{\sigma{(n+m)}}{n+m}$ hold?

Question

I would like to know more about behavior of growth rate of sum divisor function I accross this problem then :for which conditions for $n, m$ :

$$\dfrac{\sigma{(n)}}{n}\leq\dfrac{\sigma{(n+m)}}{n+m}$$ holds ?

Note:$\sigma(n)$ is the divisor function

Thank you for any help

Answer 1

Since the divisors of $n+m$ have little to do with the divisors of $n$, I doubt that you'll find a nice necessary and sufficient condition. The function $f(n) = \sigma(n)/n$ is quite irregular: here is its graph for $1 \le n \le 1000$.

One sufficient condition: if $m = k n$, then $\sigma((k+1)n) \ge (k+1) \sigma(n)$ so

$$ \dfrac{\sigma(n+m)}{n+m} = \dfrac{\sigma((k+1)n)}{(k+1)n} \ge \dfrac{\sigma(n)}{n}$$

A necessary condition is that $m+n$ is composite, since if $p$ is prime, $\sigma(p)/p = 1 + 1/p < \sigma(t)/t$ for all $t < p$.