# Calculate $\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\ \{n\sqrt{2}\} }$

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

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.

• Are you sure? {.} represents the fractional part , its not just a bracket. Commented Aug 11, 2020 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''? Commented Aug 11, 2020 at 20:37
• No, " fractional part " is defined as : $\{ x \} =x - [x]$, i.e:{3,14}=0,14. Commented Aug 11, 2020 at 20:40
• Nice one, makes sense. First time I ever see it. Thanks. Commented Aug 11, 2020 at 20:41
• The proof is well done !
– EDX
Commented Aug 11, 2020 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 $$e^{-1}$$.