# Are primes less than the sum of divisors?

I am trying to prove that

Let $p_n$ be the $n$th prime number, $\sigma (n)=\sum_{d|n}d$. Prove that $$\sigma(n) \le p_n$$

It seems obvious at first glance-to me, at least the sum of divisors of $n$ is less than the $n$th prime. In fact, it was simple to prove for numbers that are deficient or perfect, but it seems difficult to prove when $\sigma (n)>2n$.

One thing I discovered was that $$\frac{\sigma (n)}{n}=\sum_{d|n}\frac{d}{n}=\sum_{d|n}\frac{1}{d} \le H_n \Leftrightarrow \sigma(n) \le nH_n$$ Where $H_n$ is the harmonic number. I am unsure how to proceed, could anyone provide assistance? Thanks.

• $H_n$ is asymptotic to $\log n$, and $p_n$ is asymptotic to $n\log n$, so if all you have is $\sigma(n)\le nH_n$ then you are going to need error terms for those two asymptotic results in order to get anywhere. But maybe you can get a better estimate for $\sigma(n)$, as after all $n$ can't have every number up to $n$ as a factor. Apr 21 '16 at 7:21
• For example, $$\sum_{d\mid n}(1/d)\le\sum_{d\le\sqrt n}(1/d)+\sum_{d\le\sqrt n}(d/n)\le H_{\sqrt n}+1$$ and asymptotically that's $(1/2)\log n$. Apr 21 '16 at 7:25
• You can find some inequalities related to $\sigma(n)$ on this website. And if you can get hold of the book "Sandor J., Mitrinovic D.S., Crstici B. Handbook of number theory, vol.1, you can find many inequalities concerning $\sigma(n)$ in chapters III.1, III.2 and III.3. Here is Google Books link Apr 21 '16 at 7:59

G. Robin showed (1984) unconditionally that, for $n \geq 3,$ $$\sigma(n) \leq e^\gamma n \log \log n + \frac{0.6482 \; \; n}{\log \log n}$$ with the constant $0.6482...$ chosen to give equality for $n=12.$
Rosser and Schoenfeld (1962) showed that, for $n \geq 2,$ $$p_n > n \left( \log n + \log \log n - \frac{3}{2} \right)$$
• @MXYMXY I would not expect either part to have any elementary approach. You might enjoy looking at Alaoglu and Erdos, about 1944, on colossally abundant numbers. renyi.hu/~p_erdos/1944-03.pdf Also note that your $p_n / \sigma(n)$ is eventually smaller than $\sqrt n$ and any $n^{1/k}$ for any fixed $k.$ Apr 29 '16 at 19:21
• @MXYMXY thinking it over, arguments that are elementary and fairly short tell us only that both of your functions eventually lie between $n$ and $n^{1 + \delta}$ for any $\delta > 0,$ no matter how small. I do not think you can distinguish between the relevant $\log n$ and $\log \log n$ by any such easy means. Apr 29 '16 at 23:15