• $$\text{Calculate :}\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\} } . $$

  • Note:$\qquad \qquad \qquad \qquad$ Weyl's equidistributed criterion: $\quad $ The following are equivalent
    $\qquad \qquad \qquad \qquad \qquad \quad \bbox[10px,border:2px solid red] {x_n \quad \text{is equivalent modulo 1} }$ $\qquad \qquad \qquad \qquad \qquad \quad \bbox[10px,border:2px solid red] {\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f }$ $\qquad \qquad \qquad \qquad \qquad \quad \bbox[10px,border:2px solid red] {\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 }$

Im trying to approach this problem by weyl's criterion, so my thoughts so far are:

  • $\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\} } =\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\}\big)^{1/n} $

$=e^{\frac{1}{n}\log\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\}\big) }=e^{\big(\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})\big)}$

So, since $\sqrt{2}$ is irrational then its simple to prove that the sequence $x_n=\{ n\cdot \sqrt{2}\}$ is equidistributed mod1.

Let us define the continuous & 1-periodic function $f(x):=\log(x-[x])$
by the weyl's criterion we get:
$$\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})\longrightarrow\int_0^1\log(\color{black}{\underbrace{\{x\}}_{=x-[x]}})dx=\int_0^1\log(x)dx=\bigg[x\log(x)\bigg]_0^1-\int_0^1dx$$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =-1$

Hence $\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\} }=e^{-1} $.

Is there something wrong?
Also, can we find this limit with some other way?
Let me know, thank you.

  • 1
    $\begingroup$ Are you sure? {.} represents the fractional part , its not just a bracket. $\endgroup$ – John Mars Aug 11 '20 at 20:34
  • 1
    $\begingroup$ Then, no. I had never seen that notation before. Btw, what is a ''fractional part''? Don't you mean an ''integer part''? $\endgroup$ – astro Aug 11 '20 at 20:37
  • 4
    $\begingroup$ No, " fractional part " is defined as : $\{ x \} =x - [x]$, i.e:{3,14}=0,14. $\endgroup$ – John Mars Aug 11 '20 at 20:40
  • 1
    $\begingroup$ Nice one, makes sense. First time I ever see it. Thanks. $\endgroup$ – astro Aug 11 '20 at 20:41
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    $\begingroup$ The proof is well done ! $\endgroup$ – EDX Aug 11 '20 at 20:43

Since $\{ n\sqrt{2} \}$ is equidistributed modulo $2$, the limit could be rewritten as the limit of the expected value of the geometric average of $n$ uniform random variables. The integral for this would be $$\lim_{n \to \infty}\int_0^1 \int_0^1... \int_0^1 (x_1x_2...x_n)^{\frac{1}{n}} dx_n...dx_2dx_1$$

This can actually be rewritten as $$\lim_{n \to \infty}\left(\int_0^1 x^{\frac{1}{n}} dx \right)^n$$

since each $x_i$ is independent of the others. The inner integral is then equal to $\frac{n}{n+1} = 1-\frac{1}{n+1}$, so the limit is $$\lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n}$$

which is clearly $e^{-1}$.


Let $y_{n}=\{\sqrt{2}\}\{2\sqrt{2}\}\cdots \{n\sqrt{2}\}$ Assume the $\lim_{n\to+\infty} y_n $ exists , $\lim_{n\to+\infty}\sqrt[n]{y_n}=\lim_{n\to+\infty}\frac{y_{n+1}}{y_n}=\lim_{n\to+\infty}\{(n+1)\sqrt{2}\}=L$ $L$ does not exists.

  • 4
    $\begingroup$ We can't conclude from here since we have that $$\liminf_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \rightarrow \infty} a_n^{\frac{1}{n}} \leq \limsup_{n \rightarrow \infty} a_n^{\frac{1}{n}} \leq \limsup_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$ therefore when limit for the ratio exists we can conclude that it is also the limit for the n-th root otherwise this way doesn't help. $\endgroup$ – user Aug 11 '20 at 21:04

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