Upper limit of sequence involving the fractional part

Question

In the survey "The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society", the lower limit of the sequences $n\{n\sqrt 2\}, n\{n\sqrt 3\}$, where $\{\cdot\} $ is the frcational part function, are provided. I am wondering if the upper limit of such sequences is equal to $+\infty?$

Answer 1

The upper limit is $+\infty$ for the 2 sequences $(n\{n\sqrt{2}\})_n$ and $(n\{n\sqrt{3}\})_n$.

We study the first sequence (same method for the second sequence). As $\sqrt{2}$ is an irrational number, we can write it as $$\sqrt{2} = \overline{1,a_0a_1a_2...a_k...}$$ where the sequence $(a_k)_k$ contains infinite non-zero elements (because $\sqrt{2}$ is an irrational number).

Suppose these non-zero numbers are $a_{\sigma(t)}$ where $\sigma(t)$ is an strictly increasing function in $\mathbb{N}+$.

Set $n = 10^{\sigma(t)}$, then the sub-sequence $(10^{\sigma(t)}\{10^{\sigma(t)}\sqrt{2}\})_t$ tends to $+\infty$, indeed $$\begin{align} 10^{\sigma(t)}\{10^{\sigma(t)}\sqrt{2}\} &= 10^{\sigma(t)}\cdot \{1a_0a_1...a_{\sigma(t)-1} \color{red}{,}a_{\sigma(t)}a_{\sigma(t)+1}.... \}\\ &=10^{\sigma(t)}\cdot 0 \color{red}{,}a_{\sigma(t)}a_{\sigma(t)+1}....\\ &\ge 10^{\sigma(t)}\cdot 0.1 = 10^{\sigma(t)-1}\\ \end{align}$$

As $\sigma(t)$ is increasing, the sub-sequence $(10^{\sigma(t)}\{10^{\sigma(t)}\sqrt{2}\})_t$ tends to $+\infty$. Q.E.D