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

Solution 1:

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 $0$.

Solution 2:

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.