Showing $\sigma(n!) < \frac{(n+1)!}{2}$ Prove that $\sigma(n!) < \frac{(n+1)!}{2}$ for all positive integers $n$, where $n \geq 8$.
$\sigma(n)$ is sum of positive divisors of $n$.
My thought :
$n=p_1^{k_1}p_2^{k_2}...p_m^{k_m}$, where $p_1, p_2, ..., p_m$ are primes.
$\sigma(n) = \displaystyle\prod_{i=1}^m\left(\frac{p_i^{k_i+1}-1}{p_i-1}\right)$
I think this problem may be solved by using LTE.
How can we find $\sigma(n!)$ ?

The claim is equivalent to $\frac{\sigma(n!)}{n!}<\frac{n+1}{2}$ for $n \geq 8$. By inspection this holds for $n=8,9,10$, so assume $n\geq 11$ and write
\begin{align}
\frac{\sigma(n!)}{n!}&=\prod_{p\leq n}\frac{p-\frac{1}{p^{v_p(n!)}}}{p-1}\tag{1}\\
&<\prod_{p\leq n}\frac{p}{p-1}\\
&=\frac{2}{1}\frac{3}{2}\frac{5}{4}\frac{7}{6}\prod_{11\leq p\leq n}\frac{p}{p-1}\\
&\leq \frac{35}{8}\prod_{11\leq k\leq n}\frac{k}{k-1}\tag{2}\\
&=\frac{35}{8}\frac{n}{10}\tag{3}\\
&=\frac{7}{16}n\\
&<\frac{n+1}{2}.
\end{align}
Here:

*

*Definition of $\sigma$ and the fact that prime $p\mid n!$ iff $p\leq n$

*Switched from a product over primes $p$ to a product over all integers $k$ in a given range (added terms are all $>1$ so this could only increase the product)

*Product telescoped

Note: With some care the above argument can be extended to show for any $\varepsilon>0$ we have $\frac{\sigma(n!)}{n!}<\varepsilon n$ for all $n$ sufficiently large.
