# Upper limit of sequence involving the fractional part

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 $$+\infty?$$

• what is the general formula of the sequence? what is $n$ here?
– NN2
Apr 12 at 0:02
• Here $n$ is the variable of the sequence. The first sequnce is defined by $u_n=n\{n\sqrt 2\}=n(n\sqrt 2 - [n\sqrt 2]), \forall n$, the second $v_n=n\{n\sqrt 3\}=n(n\sqrt 3 - [n\sqrt 3]), \forall n$, where $[x]$ is the integer part of $x$. Apr 12 at 0:06
• $n\{ n\sqrt{3} \}$ is a second sequence? And you want to find the bounds (lower limit, upper limit) of the general sequence $n\{ n\sqrt{p} \}$ where $p\in \mathbb{N}$ when $n\to +\infty$?
– NN2
Apr 12 at 0:07
• Yes, it is a second sequence. Apr 12 at 0:09
• Just the upper limit for $p=2$ and $p=3$. The lower limit is given in the above reference. Apr 12 at 0:11

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