# On a probability density generated by the sum of reciprocols of divisors of integers

For positive integers $$n$$, define $$H(n)$$ as the sum of the reciprocals of divisors of $$n$$: $$H(n) = \sum_{k: k|n} 1/k$$ where a divisor includes both $$1$$ and $$n$$. For $$t\geq 0$$ define \begin{align} f(t) &= \limsup_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n1_{\{H(i)\geq t\}} \end{align} I would like to understand the $$f(t)$$ function (as well as perhaps its $$\liminf$$ version). In particular, I would be interested in comments or strengthened forms of the following conjecture I just made (based on a previous question linked below). My conjecture uses only a Markov inequality, but perhaps there is some stronger Gaussian approximation that can be used. Or perhaps the conjecture is false?

Conjecture: $$f(t) \leq \frac{\pi^2}{6t} \quad \forall t >0$$

Intuition: For all positive integers $$N$$ we have $$H(N) = \sum_{k=1}^{\infty} 1_{\{k|N\}}\frac{1}{k} \quad (*)$$ We can imagine that for every positive integer $$k$$, the "probability" that a "randomly selected integer $$N$$" is divisible by $$k$$ is $$1/k$$. This can be made precise by observing $$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n 1_{\{\mbox{k divides i}\}} = 1/k$$ However, for the sake of this intuitive idea, let's keep the notion of "randomly selected integer" and "probability" imprecise. So if we imagine $$N$$ as a "randomly selected large integer" then from (*) $$E[H(N)] = \sum_{k=1}^{\infty} P[\mbox{k divides N}]\frac{1}{k}= \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$$ and by using the Markov inequality (imagining that it still works in this imprecise probability world) we obtain for all $$t>0$$: $$P[H(N)\geq t] \leq \frac{E[H(N)]}{t} = \frac{\pi^2}{6t}$$ I suspect that a variation of this argument can be made rigorous by carrying out a sample path version of the Markov inequality.

This was inspired by comments on the question here:

$\lim_{n \to \infty} ​ H(a_n)=\infty$ where $H(a_n)$ is the sum of the inverses of the divisors of $a_n$

• I guess one way to make the argument rigorous is to define $H^m(n)=\sum_{k=1}^m1_{\{k|n\}}\frac{1}{k}$. Then define $N$ as a random integer uniform over $\{1, ..., m\}$ and use standard probability theory to get $P[H^m(N)\geq t] \leq E[H^m(N)]/t = \frac{1}{t}\sum_{k=1}^mP[\mbox{$k$divides$N$}]\frac{1}{k}$. Then take $m\rightarrow \infty$. Perhaps a better bound can be obtained by using second moments and Chebyshev. Commented Jun 12 at 23:21
• Just to be clear, the function $H(n)$ is the abundancy of the integer $n$? More commonly expressed as $\frac{\sigma(n)}{n}$. Commented Jun 13 at 7:26
• @Servaes We have $H(1)=1$, $H(2)=1 + 1/2$, $H(3) = 1+1/3$, $H(4)=1+1/2+1/4$, $H(5)=1+1/5$, and so on. The Sungjin answer relates this to a $\sigma(n)/n$ function that I am not familiar with. I formulated the problem by trying to cast the linked question into a precise statement. The linked question seems to suggest that $f(t)\rightarrow 1$, but I realized the above informal argument suggested it decays like $O(1/t)$. It turns out that my $\pi^2/(6t)$ conjecture is correct (even in a nonasymptotic sense), and also $\frac{1}{n}\sum_{i=1}^nH(i)\rightarrow \pi^2/6$. Commented Jun 13 at 13:17
• Thanks for the clarification. Here $\sigma(n)$ denotes the sum of all positive divisors of $n$, so indeed $H(n)=\tfrac{\sigma(n)}{n}$. The function $\sigma(n)$ has been studied very extensively, and any similar questions you might have are likely easily answered with known asymptotics of $\sigma(n)$ in a fairly straightforward way. Either way, nice question and glad your question was answered! Commented Jun 13 at 17:23

## 2 Answers

We have $$H(n)=\sum_{k=1}^{\infty} 1_{\{k|n\}}\frac{1}{k} =\sum_{k|n}\frac kn$$ by reversing the role of $$k$$ and $$n/k$$. Then a well-known sum-of-divisor fuction appears. From the above, $$H(n)=\frac{\sigma(n)}n.$$ Thus, the sum you are looking for $$A(t)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n1_{\{H(i)\geq t\}}$$ is an asymptotic density of positive integers with $$\sigma(i)/i \geq t$$. Such asymptotic density is known to exist and continuous in the variable $$t>0$$.

Andreas Weingartner has number of results in this direction. I will point you to one of his results here: PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 135, Number 9, September 2007, Pages 2677–2681 S 0002-9939(07)08771-0

His Theorem 1 states that as $$t\rightarrow\infty$$, $$A(t)=\exp\left( -e^{te^{-\gamma}}\right)(1+O(t^{-2}))$$ where $$\gamma=0.5772...$$ is Euler's constant.

• Thanks! From the paper you include, it looks like $f(t)$ decays much faster than $O(1/t)$. I have added an answer that shows the $\pi^6/(6t)$ bound holds nonasymptotically for all $n$ (just refining the Markov inequality argument in my question). Commented Jun 13 at 1:13

This just makes my Markov inequality argument rigorous and stronger.

Define $$\mathbb{N}$$ as the set of positive integers. Let $$g:\mathbb{N}\rightarrow [0, \infty)$$ be a nonnegative function. For $$n \in \mathbb{N}$$ define $$G(n) = \sum_{k:k|n} g(k)$$ The special case $$g(k)=1/k$$ for all $$k$$ yields $$G(n)=H(n)$$.

Claim 1: For all $$n \in\mathbb{N}$$ and $$t>0$$ we have $$\frac{1}{n}\sum_{i=1}^n1_{\{G(i)\geq t\}} \leq \frac{1}{t}\sum_{k=1}^{\infty}\frac{g(k)}{k}$$ The special case $$g(k)=1/k$$ for all $$k$$ yields $$\frac{1}{n}\sum_{i=1}^n1_{\{G(i)\geq t\}} \leq \frac{\pi^2}{6t}$$.

Claim 2: If $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^ng(i)=0$$ then $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^nG(i)=\sum_{k=1}^{\infty}\frac{g(k)}{k}$$ The special case $$g(k)=1/k$$ for all $$k$$ yields $$\frac{1}{n}\sum_{i=1}^nG(i)\rightarrow\pi^2/6$$.

Proof of Claim 1: Fix $$n \in \mathbb{N}$$ and $$t>0$$. For each $$i \in \{1, ..., n\}$$ we have $$G(i) = \sum_{k=1}^n 1_{\{k|i\}} g(k) \quad (Eq \: 1)$$ Define a discrete random variable $$N \sim \mbox{Unif}\{1, ..., n\}$$. By the law of total probability $$P[G(N)\geq t] = \sum_{i=1}^nP[G(i)\geq t | N=i]P[N=i] = \frac{1}{n}\sum_{i=1}^n1_{\{G(i)\geq t\}}$$ On the other hand, since $$G(N)$$ is a nonnegative random variable, the Markov inequality gives \begin{align} P[G(N)\geq t] &\leq \frac{E[G(N)]}{t} \\ &\overset{(a)}{=}\frac{1}{t}E\left[\sum_{k=1}^n1_{\{k|N\}}g(k)\right]\\ &= \frac{1}{t}\sum_{k=1}^nP[\mbox{k divides N}]g(k) \end{align} where equality (a) holds by Eq (1). Now observe for all $$k \in \{1, ..., n\}$$ that $$P[\mbox{k divides N}] =\frac{\lfloor n/k\rfloor}{n}\leq 1/k$$ $$\Box$$

Proof of Claim 2: Fix $$n \in \mathbb{N}$$. Define random variable $$N \sim \mbox{Unif}\{1, ..., n\}$$. Then $$E[G(N)] = \frac{1}{n}\sum_{i=1}^nG(i)$$ On the other hand, from Eq (1) we have $$G(N) = \sum_{k=1}^n1_{\{k|N\}}g(k)$$ Taking expectations of both sides gives $$E[G(N)] = \sum_{k=1}^nP[\mbox{k divides N}]g(k)$$ Also $$P[\mbox{k divides N}] = \frac{\lfloor n/k\rfloor}{n} \in \left[\frac{1}{k}-\frac{1}{n}, \frac{1}{k}\right]$$ and so $$\sum_{k=1}^n\frac{g(k)}{k} - \sum_{k=1}^n\frac{g(k)}{n} \leq E[G(N)]\leq \sum_{k=1}^n\frac{g(k)}{k}$$ Taking $$n\rightarrow\infty$$ gives the result. $$\Box$$