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I want to show that $$\limsup_{n \rightarrow \infty} \frac{\sigma(n)}{n \log \log n} = e^{\gamma}.$$ where $\sigma(n):=$ sum of positive divisor function and $\gamma$ is the Euler constant (the one appear in Merten's estimates $(3)$ https://en.wikipedia.org/wiki/Mertens%27_theorems)

I try to use Merten estimate since it appear $\gamma$ in the limit valaue. Since $$\sigma(p_1^{a_1}...p_k^{a_k})= \frac{\prod_{i=1}^k (p_i^{a_i} -1/p_i)}{\prod_{i=1}^k (1-1/p_i)}.$$ The reciprocal might be modified to use Merten reciprocal form $$\prod_{p \leq x; p\ \ prime} (1-1/p)^{-1} = e^{\gamma} \log x (1 + O(1/\log x)).$$ I guess that $\log \log$ thing, and $\log$ and error term in Merten should be somehow cancel, but really not sure why.

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Let $N_x = \prod_{p < x} p$ the primorial. By the Mertens theorems we have $$\frac{\sigma(N_x)}{N_x} = \prod_{p < x} (1+\frac{1}{p} )= e^{\gamma+\ln \ln x+o(1)}$$ Now $\ln N_x = \sum_{p < x} \ln p $ so that by Chebyshev's work (*) $x/2 < \ln N_x <2x$ and $\ln x =\ln \ln N_x+ \mathcal{O}(1)$, $\ln \ln x =\ln \ln \ln N_x+ o(1)$ and we get $$\frac{\sigma(N_x)}{N_x} = e^{\gamma+\ln \ln \ln N_x+o(1)} = e^{\gamma+o(1)}\ln \ln N_x$$ and finish with $$\lim \sup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n} = \lim \sup_{x \to \infty} \frac{\sigma(N_x)}{N_x \ln \ln N_x} = e^{\gamma}$$


The Mertens theorem we need is proved there, but I can try to make it shorter :

Let $\Lambda(p^k) = \ln p$ if $p$ is prime, $\Lambda(n) = 0$ otherwise. By a simple sieving/combinatoric we have $x!= \prod_{p^k \le x} p^{ \lfloor x /p^k \rfloor}$ so that $$\sum_{n \le x} \lfloor x /n\rfloor \Lambda(n) = \sum_{p^k \le x} \lfloor x /p^k \rfloor \ln p = \ln x!= x \ln x+ \mathcal{O}(x)$$ by Stirling's approximation. Hence $$\sum_{n \le x} \frac{\Lambda(n)}{n} = \frac{1}{x} \sum_{n \le x} \Lambda(n)(\lfloor x /n\rfloor+\mathcal{O}(1)) = \ln x+ \mathcal{O}(1)$$ where we used that $\sum_{n \le x}\Lambda(n) = \mathcal{O}(x)$

Finally with $\ln x = \mathcal{O}(1) +\sum_{n \le x} \frac{1}{n}$ we get

$$\sum_{n < x} \frac{\Lambda(n)-1}{n} = \mathcal{O}(1)$$ Summing by parts (mistakes here) $$\mathcal{O}(1) = \sum_{n =2}^x \frac{\Lambda(n)-1}{n \ln n}\ln n = \ln x\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n}+\sum_{2 \le k < x}(\ln k-\ln (k+1))\sum_{2\le n < k} \frac{\Lambda(n)-1}{n \ln n} = \ln x\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n} + \sum_{2 \le k < x} \mathcal{O}(1/k)$$ so that $\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n} = \mathcal{O}(1/\ln n)$ so we can refine $\sum_{2 \le k < x} \mathcal{O}(1/k)$ to $\sum_{2 \le k < x} \mathcal{O}(\frac{1}{k \ln k}) = \mathcal{O}(\ln \ln x)$ and get $$\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n} = o(1)$$ Finally $\sum_{n \le x} \frac{\Lambda(n)}{n \ln n} = o(1)-\sum_{p \le x} \ln(1-\frac{1}{p})$ and $$\sum_{p \le x}\ln(1+\frac{1}{p}) = o(1)+\sum_{p \le x} (\ln(1-\frac{1}{p^2})-\ln(1-\frac{1}{p})) = - \ln \zeta(2)+o(1)+\sum_{n \le x} \frac{\Lambda(n)}{n \ln n} =- \ln \zeta(2)+o(1)+\sum_{2 \le n \le x} \frac{1}{n \ln n} =\gamma+o(1)+\ln \ln x$$ where I used $\sum_{2 \le n \le x} \frac{1}{n \ln n} = \ln \ln(x)+\gamma-\zeta(2)+o(1)$, no reference for the $\gamma-\zeta(2)$ constant

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In order to exploit Mertens' theorem, you just have to show that $\frac{\sigma(n)}{n}$ is maximized when $n$ is a primorial. By the PNT, $\sum_{p\leq x}\log p = x+o(x)$, hence if we take $n$ as $\prod_{p\leq x}p$, we have

$$ \frac{\sigma(n)}{n\log\log n}\approx \frac{1}{\log x}\prod_{p\leq x}\left(1+\frac{1}{p}\right)\approx\frac{1}{\log x}\prod_{p\leq x}\left(1-\frac{1}{p}\right)^{-1}.$$

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  • $\begingroup$ See there you also need a few more steps, but the PNT is enough (though not necessary) $\endgroup$ – reuns Oct 2 '16 at 21:03
  • $\begingroup$ @user1952009: I agree your answer is much more accurate than mine. I just wanted to give the OP the main ideas here, that is the reason for using $\approx$ instead of more accurate (asymptotic) bounds. $\endgroup$ – Jack D'Aurizio Oct 2 '16 at 21:07
  • $\begingroup$ I gave the link because it shows where $\gamma$ comes from (without details) $\endgroup$ – reuns Oct 2 '16 at 21:09
  • $\begingroup$ @user1952009: and that is a fine idea. Thanks. $\endgroup$ – Jack D'Aurizio Oct 2 '16 at 21:11
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    $\begingroup$ @JackD'Aurizio if you are interested, I posted a sketch-proof that should work without the PNT $\endgroup$ – reuns Oct 2 '16 at 22:07

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