# $\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\} }$

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

• 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.

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