2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ $H_n$ 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. $\endgroup$ Apr 21 '16 at 7:21
  • $\begingroup$ 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 $(1/2)\log n$. $\endgroup$ Apr 21 '16 at 7:25
  • $\begingroup$ 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 $\endgroup$ Apr 21 '16 at 7:59
1
$\begingroup$

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 $n \geq 2,$ $$ p_n > n \left( \log n + \log \log n - \frac{3}{2} \right) $$

$\endgroup$
3
  • $\begingroup$ Would there be some elementary way to show this? $\endgroup$
    – S.C.B.
    Apr 29 '16 at 11:30
  • $\begingroup$ @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 $k.$ $\endgroup$
    – Will Jagy
    Apr 29 '16 at 19:21
  • $\begingroup$ @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. $\endgroup$
    – Will Jagy
    Apr 29 '16 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.