I am trying to show that $\frac{\sigma(m!)}{m!}>ln(m)$, where $m$ is a positive integer, and I can't seem to get anywhere in my proof. Where $\sigma(n)$ is the sum of the divisors of $n$.
I've tried to write $\sigma(m!)$ after taking an arbitrary prime power decomposition for m! and can't seem to get anywhere.
I'm personally thinking this has something to do with Merten's function since: $log(x)=\sum_{n\leq x} \frac{\Lambda(n)}{n}-O(1)$ with $\Lambda(n)=log(p)$ if $n$ is a prime power and 0 otherwise. I've tried expanding this function and playing around with it but still can't make headway relating this back to $\sigma$.
Any suggestions or hints would be greatly appreciated!
Best Answer
Hint: Note that the numbers $m!, \dfrac{m!}{2}, \dfrac{m!}{3},\ldots,\dfrac{m!}{m-1}, \dfrac{m!}{m}$ are all distinct divisiors of $m!$. So $$\sigma(m!) \ge m!+\dfrac{m!}{2}+ \dfrac{m!}{3}+\ldots+\dfrac{m!}{m-1}+\dfrac{m!}{m}.$$ See if you can use that along with elementary bounds on the harmonic numbers to solve the problem.