# there exist infinite many $n\in\mathbb{N}$ such that $S_n-[S_n]<\frac{1}{n^2}$

recent conjecture ：Let $S_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$, where $n$ is a positive integer. Prove that ：there exist infinite many $n\in\mathbb{N^{+}}$ such that $$S_n-[S_n]<\dfrac{1}{n^2}$$ where $[x]$ represents the largest integer not exceeding $x$.

previous problem:How to prove that $a<S_n-[S_n]<b$ infinitely often

• Can you find any example with $n \gt 1$? Having looked at this for small $n$, I would guess there are probably no other examples. – Henry Dec 18 '16 at 1:33
• Heuristically this should happen a finite number of times. Consider the values $n_i$ where $\lfloor H_{n_i}\rfloor$ jumps from $i$ to $i+1$. Since $H_i\approx\ln i$, we have $n_i\approx e^i$ (this is of course a very rough approximation, but that's all we need for this heuristic). Now, at $n_i$ it's reasonable to assume (heuristically) that the fractional part $\{H_{n_i}\}$ is equidistributed in $(0, \frac1{n_i})$; this means that the probability it's less than $1/(n_i^2)$ is $\frac1{n_i}$. But then the total number of 'hits' is $\sum_i(n_i^{-1})\approx\sum_i e^{-i}$, which converges. – Steven Stadnicki Dec 23 '16 at 2:07
• @alphacapture There's nothing specific about $\{H_{n_i}\}$ here - the heuristic is that we've added $1/n_i$ to $H_{n_i-1}$ to get a number that 'rolls over' into the next integer, so we know that $\{H_{n_i}\}\geq 0$ (by definition) and $\{H_{n_i}\}\leq 1/n_i$ (because otherwise the rollover point would've been earlier); the heuristic is just in the assertion it's equidistributed within its possible range (because we have no specific reason to believe otherwise). – Steven Stadnicki Dec 23 '16 at 23:25
• Formally, we can say that the set $\{\alpha \in [0,1] : (S_n+\alpha)-[S_n+\alpha] < 1/n^2 \ \text{for infinitely many} \ n \in \mathbb{N}^+\}$ has measure $0$. But showing that $0$ does not belong to this set doesn't appear to be trivial. – JimmyK4542 Dec 24 '16 at 7:42
• You should probably add to your post that you have now asked the same question on [MathOverflow](there exist infinite many $n\in\mathbb{N}$ such that $S_n-[S_n]<\frac{1}{n^2}$). See this answer and other discussions about cross-posting. – Martin Sleziak Jan 15 '17 at 11:42

## 2 Answers

This is a heuristic argument. It does not show that there are not infinitely many occurrences, just that without some unforeseen conditions, the probability of infinitely many occurrences is zero.

The Euler-Maclaurin Sum Formula says that $$H_n=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}+O\!\left(\frac1{n^6}\right)\tag{1}$$ Subtracting $\gamma$ and exponentiating gives \begin{align} e^{H_n-\gamma} &=ne^{\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}}+O\!\left(\frac1{n^5}\right)\\ &=\left(\frac1n-\frac1{2n^2}+\frac5{24n^3}-\frac1{16n^4}+\frac{47}{5760n^5}+\frac1{2304n^6}\right)^{-1}+O\!\left(\frac1{n^5}\right)\tag{2} \end{align} Inversion of the power series $y=x-\frac{x^2}2+\frac{5x^3}{24}-\frac{x^4}{16}+\frac{47x^5}{5760}+\frac{x^6}{2304}$ says $$x=y+\frac{y^2}2+\frac{7y^3}{24}+\frac{y^4}6+\frac{523y^5}{5760}+\frac{y^6}{20}+O\!\left(y^7\right)\tag{3}$$ where $x=\frac1n$ and $y=e^{\gamma-H_n}$. Taking the reciprocal of $(3)$ yields $$n=e^{H_n-\gamma}-\frac12-\frac1{24}e^{\gamma-H_n}+\frac3{640}e^{3\gamma-3H_n}+O\!\left(\frac1{n^5}\right)\tag{4}$$ To find an $n$ so that the fractional part of $H_n$ is less than $\frac1{n^2}$, we need to find an $H_n$ so that the fractional part of $n$ is greater than $1-\frac1n$. $$\begin{array}{r|r} H_n&n\\\hline 1&1.000000000000\\ 2&3.638675849525\\ 3&10.773523676598\\ 4&30.153290055642\\ 5&82.827475640215\\ 6&226.008738099299\\ 7&615.215019121592\\ 8&1673.187107043897\\ 9&4549.053308117194 \end{array}\tag{5}$$ Without a reason that the fractional part of $(4)$ not be uniformly distributed, the probability that the fractional part of $H_n$ is less than $\frac1{n^2}$ would be $$\frac1n\approx e^{\gamma-H_n}\tag{6}$$ and since \begin{align} \sum_{H_n=1}^\infty e^{\gamma-H_n} &=\frac{e^\gamma}{e-1}\\ &\lt\infty\tag{7} \end{align} Borel-Cantelli says that the probability that there are infinitely many $n$ so that the fractional part of $H_n$ is less than $\frac1{n^2}$ is $0$.

• How much chance that the Pythagorean Theorem is true? I guess someone can give a probabilistic "proof" that it is zero too. – Han de Bruijn Jan 12 '17 at 20:13
• I explicitly said that this was not a proof, that it was only heuristic. If someone posts a proof, this would serve to enhance the wonder because the odds are against it. If nothing else, the inverse of the harmonic numbers was worth the exercise for me. – robjohn Jan 12 '17 at 20:20
• I hope that some of the results that I have gotten, while they don't settle the question, are helpful to someone. – robjohn Jan 12 '17 at 20:40
• Sure Rob, but I only wanted to point out that one has to be careful with transfer of probabilistic arguments to environments where they do not really belong. – Han de Bruijn Jan 12 '17 at 20:59
• @HandeBruijn Out of curiosity, how do you (propose to) cast the Pythagorean Theorem in such a form as to be able to apply probabilistic heuristics to it? – Steven Stadnicki Jan 20 '17 at 22:33

Sorry for posting as answer but I think it would be more convenient here. Firstly, I am offering to change the title of the question to ''There is no $n$ such ... ''. It might be more attractive. Secondly, I have the following generalized conjecture:

Let $M$ be the set of all $n$ such that $$H_n - \lfloor{H_n\rfloor} < \frac{1}{n^{1+\epsilon}}.$$ Then $$\forall\epsilon>0 : |M| = \bar\eta(\epsilon) < \infty.$$

Also I suspect that $\bar\eta$ monotonically decreasing function on $(0,1]$. Below I presented possible values of $\bar\eta$ ('possible' because I am not able to check all $n$'s). Actually, I was checking only up to $10^5$ first $n$'s. Next, I tried all $n$'s up to $10^6$. Note that I assume $H_n$ starts with $\frac12$, not with 1.

• Obviously false. $S_n-[S_n]<\frac1n$ holds about every n of $S_n-[S_n]<m<S_{n+1}-[S_{n+1}]$. – Takahiro Waki Jan 21 '17 at 6:14
• @TakahiroWaki $\epsilon > 0$. – LRDPRDX Jan 21 '17 at 8:24
• However even the case of $\frac1{n^2}$, now anyone can't say there doesn't exist one solution. As I say, we need to search up to $10^{100}$. Although you claim this conjecture holds even If n can be reduce consectively, there is not such a logic in your answer. Of course to find such a logic is more hard than solving this conjecture. – Takahiro Waki Jan 21 '17 at 9:36
• So. What logic is behind the statement of this question? – LRDPRDX Jan 21 '17 at 9:52
• Do you claim $ε$ is $0.8$? – Takahiro Waki Jan 21 '17 at 10:28