How to show $\frac{\sigma(m!)}{m!}>ln(m)$?

Question

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!

Answer 1

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.

Answer 2

If you are interested in the harder question of the asymptotics for $\sigma(n!)/n!$ then one can prove that it essentially behaves as $$ a_n = \mathrm e ^\gamma \log n$$ where $\gamma $ is the Euler--Mascheroni constant. By this I mean that $$\lim_{n\to \infty} \frac{\sigma(n!)}{n! a_n} =1 .$$ For an integer $m $ let $\alpha_p(m)$ be the largest power of a prime $p$ that divides $m$. Then we have $$ \frac{\sigma (m)}{m} = \sum_{d\mid m} \frac{d}{m}=\sum_{d\mid m } \frac{1}{d}=\prod_{p\mid m } \left( 1+ \frac{1}{p}+\ldots + \frac{1}{p^{\alpha_p(m) }} \right) .$$ For an upper bound we have $$ \frac{\sigma (n!)}{n!} \leq \prod_{p\leq n } \frac{1}{1-1/p} ,$$ which, when divided by $a_n $ tends to $1$ by Mertens' third theorem. The lower bound is a bit harder. First, $$ 1+ \frac{1}{p}+\ldots + \frac{1}{p^\alpha} =\frac{1}{1-1/p} \left(1 -\frac{1}{ p^{\alpha +1 } } \right) $$ so that $$ \frac{\sigma(n!)}{n! a_n } \geq \frac{\prod_{p\leq n } (1-1/p)}{a_n} \prod_{p\leq n } \left(1 -\frac{1}{ p^{\alpha_p(n) +1 } } \right).$$ The first fraction in the right side tends to $1$ again by Mertens' third theorem. Hence, we only have to prove that $$ \lim_{n\to \infty } \prod_{p\leq n } \left(1 -\frac{1}{ p^{\alpha_p(n) +1 } } \right) =1 .$$ Using the inequality $ (1-x) \geq e^{-2x}$ which is valid for all $0<x\leq 1/2$ we obtain $$ \prod_{p\leq n } \left(1 -\frac{1}{ p^{\alpha_p(n) +1 } } \right) \geq \exp\left(-2 \sum_{p\leq n } \frac{1}{ p^{\alpha_p(n) +1 } } \right) $$ and hence it suffices to prove $$ \sum_{p\leq n } \frac{1}{ p^{\alpha_p(n) +1 } } \to 0 .$$ For this, one has to use that $\alpha_p(n!)$ is always kind of large for every small $p$. By Legendre's lemma $$ \alpha_p(n!) = \sum_{k=1}^\infty \left[ \frac{n}{p^k} \right] $$ we get that $a_p(n)\geq 1$ for $ n^{1/2 }<p \leq n $. Hence, $$ \sum_{\sqrt{n} < p\leq n } \frac{1}{ p^{\alpha_p(n) +1 } } =\sum_{\sqrt{n} < p\leq n } \frac{1}{ p^2 } \ll \frac{1}{\sqrt{n} }.$$ For the remaining primes $p$ we have $p\leq n^{1/2 } \leq n/2$ hence $[n/p] \geq n/(2p) $ hence $$ \alpha_p(n) \geq \frac{n}{2 p} \geq \frac{n}{2 n^{1/2} } =\frac{ \sqrt{n} }{2} \Rightarrow \frac{1}{p^{1+\alpha_p(n) } } \leq \frac{1}{2^{1+\alpha_p(n) } } \leq \frac{1}{2^{ \sqrt{n } /2 } }\Rightarrow \sum_{p\leq \sqrt{n} } p^{-1-\alpha_p(n)} \leq \frac{\sqrt{n} }{2^{ \sqrt{n } /2 } } \to 0 .$$ This concludes the proof. With a bit more details the same argument gives an explicit rate of convergence, namely $$ \frac{\sigma (n!)}{n!} =\mathrm e ^\gamma (\log n) +O\left(\frac{1}{n }\right).$$