Question about Certain Upper Bounds of the Sum of Divisors Function

Question

It has been proven that $$\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n}=e^\gamma$$ where $\sigma(n)$ is the sum of divisors and $\gamma$ is Euler's constant ($0.5772...$). The way I understand it is that for every $\epsilon$, there is some $b$, so that when $n>b$, $(e^\gamma + \epsilon)n \ln\ln n>\sigma(n)$. Is there a way to find such smallest $b$ for every $\epsilon$ to make above inequality true (that is, for any smaller $b$, the inequality wouldn't hold)? If there is not a way to find the exact smallest $b$, is there at least a way to find a "tight" estimate/upper bound (any smaller estimate/upper bound wouldn't work) on such smallest $b$?

Answer 1

We have the following (unconditional) inequality due to Robin: $$\sigma(n)<e^\gamma n\log\log n+\frac{0.6483 n}{\log\log n}$$ for $n\geq 3$. Therefore, to have the inequality you ask it's enough to have $$e^\gamma n\log\log n+\frac{0.6483 n}{\log\log n}\leq(e^\gamma+\varepsilon)n\log\log n.$$ Elementary manipulations show that this is equivalent to $$n\geq e^{e^{\sqrt{0.6483/\varepsilon}}},$$ so we may take the right-hand side as the upper bound for $b$, provided it's at least $3$.

There are better inequalities known nowadays, for instance the following due to Axler (A new upper bound for the sum of divisors function) gives $$\sigma(n)<e^\gamma n\log\log n+\frac{0.1209 n}{(\log\log n)^2},$$ for $n>2520$, and we can similarly derive an upper bound for $b$: $$b\leq e^{e^{\sqrt[3]{0.1209/\varepsilon}}}$$ if the right-hand side is at least $2520$.

As for finding the exact value of $b$, this already shows that this is in principle algorithmically possible, since we just have to find the largest $n$ smaller than the given bound for a given $\varepsilon$.

If we admit conjectural bounds, by results of Robin under RH the inequality holds for $\varepsilon=0$ for $n>5040$, and therefore $b\leq 5040$ for all $\varepsilon$, and for all small enough values of $\varepsilon$ this is exact. Therefore under RH, $b=5040$ for small enough $\varepsilon$.

Answer 2

• The definition of $\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n}=e^\gamma$ is $$f(n) = \sup_{m \ge n} \frac{\sigma(m)}{m \ln \ln m}, \qquad \lim_{n \to \infty} f(n) = e^{-\gamma}$$

• $\forall \epsilon>0, \exists b,\forall n \ge b,| f(n) -e^{-\gamma}|<\epsilon$ is the definition of $\lim_{n \to \infty} f(n) = e^{-\gamma}$.

• $\lim_{n \to \infty} f(n) = e^{-\gamma}$ follows from Mertens theorems.

• Robin Nicolas Lagarias proved that the RH holds iff $\forall n,f(n) = e^{-\gamma}$.

• Thus $f(n)$ is computable iff the RH is (dis)provable. If it is computable then for a fixed $\epsilon$ the smallest $b$ is computable.