Let $\sigma$ be the sum-of-divisors function defined by $$ \sigma(n) = \sum_{d \mid n} d. $$ Is there an explicit lower bound for $\sigma(n)/n$ in the style of the lower bound $$ \phi(n) > \dfrac{n}{e^\gamma \log \log n + \frac{3}{\log \log n}} $$ for $n > 2$, where $\phi$ is the Euler's totient function and $\gamma$ is the Euler-Mascheroni constant? I'm aware that the inequality $$ \dfrac{6}{\pi^2} < \dfrac{\phi(n) \sigma(n)}{n^2} < 1, $$ for $n > 1$, is known. But can we get a better lower bound than $\sigma(n)/n > 6n/\pi^2\phi(n)$?
Asked
Active
Viewed 1,980 times
2
-
Interesting upper bounds, yes. Lower bounds, no. Whenever $n$ is prime, $$ \frac{\sigma(n)}{n} = 1 + \frac{1}{n}. $$ – Will Jagy May 28 '14 at 00:57
-
upper bounds, short version at http://math.stackexchange.com/questions/808168/how-to-prove-dn2n-log-log-n/808178#808178 – Will Jagy May 28 '14 at 01:03
2 Answers
2
Whenever n is prime, Sigma(n)=n+1
Whenever n>1 is composite Sigma (n)> n+ (n)^0.5
The proof is found at: http://www.americanriver.com/mathematics/elementary_number_theory.html under chaper 6.1 " the functions tau and sigma" : task 10.c)
1
You can't do better if you want a lower bound that holds for all $n$. If you consider the sequence of integers which are products of the consecutive primes, i.e. $$2, \;2\cdot3, \;2\cdot3\cdot5, \;2\cdot3\cdot5\cdot7,\,\dots$$ you can see that your bound is sharp.
For better bounds on subsets of the positive integers, you might want to look at the following this paper by Medryk.
John M
- 7,508
-
I guess this is what he wanted. Meanwhile, take a look at http://math.stackexchange.com/questions/323144/eulers-phi-function-worst-case/323229#323229 as the Nicolas criterion deserves to be better known. – Will Jagy May 28 '14 at 02:06
-