Peter 25.05.2007 03:25 Show that for any positive integer $n$, \[\frac{\sigma(n!)}{n!}\ge \sum_{k=1}^{n}\frac{1}{k}.\]
Peter 22.07.2007 06:13 Robert Gerbicz wrote: For all $ 1\leq k\leq n$ integers $ \frac{n!}{k}$ are distinct positive divisors of $ n!$, so $ \sigma(n!)\geq \sum_{k=1}^{n}\frac{n!}{k}$ Dividing this inequality by $ n!$ we can get the solution.