Which of these two expressions (related to primes) is greater: $\frac{\sigma{(n)}}{n}$ or $\sum_{p \le n}{\frac{1}{p}}$?

Question

Would it be $\frac{\sigma{(n)}}{n}$ or $\sum_{p \le n}{\frac{1}{p}}$?

Why?

Thank you!

Answer 1

Neither. Couple calculations show:

$n=6$: $$\frac{\sigma{(6)}}{6}= \frac{1+2+3+6}{6} = 2 > \frac{31}{30} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \sum_{p \le 6}{\frac{1}{p}}$$

$n=7$: $$\frac{\sigma{(7)}}{7}= \frac{1+7}{7} = \frac{8}{7} < \frac{247}{210} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} = \sum_{p \le 7}{\frac{1}{p}}$$

And just for fun $n=101$ you have $$\frac{\sigma{(101)}}{101} = \frac{102}{101} < \frac{422113843906354093775418512493046577809}{232862364358497360900063316880507363070} = \sum_{p \le 101}{\frac{1}{p}}$$

but for $n=102$ you have $$\frac{\sigma{(102)}}{102} = \frac{36}{17} > \frac{422113843906354093775418512493046577809}{232862364358497360900063316880507363070} = \sum_{p \le 102}{\frac{1}{p}}$$