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 $+\infty?$

  • $\begingroup$ what is the general formula of the sequence? what is $n$ here? $\endgroup$
    – NN2
    Apr 12 at 0:02
  • $\begingroup$ 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 $x$. $\endgroup$ Apr 12 at 0:06
  • $\begingroup$ $n\{ n\sqrt{3} \}$ 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 $n\to +\infty$? $\endgroup$
    – NN2
    Apr 12 at 0:07
  • $\begingroup$ Yes, it is a second sequence. $\endgroup$ Apr 12 at 0:09
  • $\begingroup$ Just the upper limit for $p=2$ and $p=3$. The lower limit is given in the above reference. $\endgroup$ Apr 12 at 0:11

1 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

  • 1
    $\begingroup$ The argument is very clear. Thanks $\endgroup$ Apr 12 at 11:19

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