$$\text{Calculate :}\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} } . $$
Note:
Weyl's equidistributed criterion. The following are equivalent:
$$x_n\quad\text{is equidistributed modulo 1}$$
$$\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f $$
$$\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 $$
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}=\\
&=\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:
$$\begin{align*}
\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})&\longrightarrow\int_0^1\log(\color{black}{\underbrace{\{x\}}_{=x-[x]}})dx\\[5pt]
&=\int_0^1\log(x)dx\\[5pt]
&=\bigg[x\log(x)\bigg]_0^1-\int_0^1dx\\[5pt]
&=-1\\[5pt]
\end{align*}$$
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.
Best Answer
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 $e^{-1}$.