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


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

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

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


Background:
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}\\ &=e^{\frac{1}{n}\log\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big) }\\ &=e^{\big(\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})\big)}\\ \end{align}

So, since $\sqrt{2}$ is irrational then its 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:

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

Hence

$$\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }=e^{-1} $$ Is there something wrong? Also, can we find this limit with some other way?

Let me know, thank you.

Jessie
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2 Answers2

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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 $e^{-1}$.

Varun Vejalla
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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.

ratatuy
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    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. – user Aug 11 '20 at 21:04