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

Weyl's equidistributed criterion. The following are equivalent: $$x_n\quad\text{is equidistributed modulo 1}$$

$$\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f $$

$$\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 $$

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

\begin{align} &\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }\\ &=\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big)^{1/n}=\\ &=\exp\left(\frac{1}{n}\log\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big) \right)=\\ &=\exp\left(\frac{1}{n}\sum_{k=1}^n \log\big(\{k\sqrt{2}\}\big)\right)\\ \end{align}

So, since $\sqrt{2}$ is irrational then it's simple to prove that the sequence $x_n=\{ n\cdot \sqrt{2}\}$ is equidistributed $\text{mod}\ 1$. Let us define the continuous & $1$-periodic function $f(x):=\log(x-[x])$
by the weyl's criterion we get:

$$\begin{align*} \frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})&\longrightarrow\int_0^1\log(\color{black}{\underbrace{\{x\}}_{=x-[x]}})dx\\[5pt] &=\int_0^1\log(x)dx\\[5pt] &=\bigg[x\log(x)\bigg]_0^1-\int_0^1dx\\[5pt] &=-1\\[5pt] \end{align*}$$


$$\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }=e^{-1} $$ 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. $\endgroup$ Aug 11, 2020 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, 2020 at 20:37
  • 4
    $\begingroup$ No, " fractional part " is defined as : $\{ x \} =x - [x]$, i.e:{3,14}=0,14. $\endgroup$ Aug 11, 2020 at 20:40
  • 1
    $\begingroup$ Nice one, makes sense. First time I ever see it. Thanks. $\endgroup$
    – astro
    Aug 11, 2020 at 20:41
  • 6
    $\begingroup$ The proof is well done ! $\endgroup$
    – EDX
    Aug 11, 2020 at 20:43

1 Answer 1


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 $e^{-1}$.

  • $\begingroup$ typo I think: the sum is equidistributed "modulo 1" $\endgroup$
    – Thomas
    Nov 15, 2022 at 15:47
  • $\begingroup$ question: can you please show that that sequence is "equidistributed modulo 1" and formalize that statement ? $\endgroup$
    – Thomas
    Nov 15, 2022 at 15:49
  • $\begingroup$ How do you rigorously justify the interchange between an equidistributed sequence and a uniform random variable? It's not clear to me what clever choice of probability space justifies this $\endgroup$
    – FShrike
    Nov 16, 2022 at 16:31

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