Calculate limit with floor function

Question

How can I proceed to find the following limit:

$$\lim_{n \rightarrow \infty }(n\sqrt 2-\lfloor n\sqrt2 \rfloor) $$

where $n$ is a natural number. Please if there is no limit, would you provide a proof.

Answer 1

Let $a_n=n\sqrt 2-\lfloor n\sqrt2 \rfloor$. Note that $a_n>0$ for all $n\in\mathbb{Z^+}$.

Suppose the sequence converges to a limit, $L$, and choose $\epsilon=0.01$, then we have for all $n\ge N$ (for some choice of $N\in\mathbb{N}$),

$$0.01<a_n-L<0.01$$

Then

$$\begin{align} a_{n+1}-a_n&=((n+1)\sqrt 2-\lfloor (n+1)\sqrt2 \rfloor)-(n\sqrt 2-\lfloor n\sqrt2 \rfloor) \\[2ex] &=\sqrt 2+\lfloor n\sqrt2 \rfloor-\lfloor (n+1)\sqrt2 \rfloor \end{align}$$ The quantity on the right-hand side is equal to either $\sqrt2-1$ or $\sqrt2-2$.

In the first case

$$\begin{align} a_{n+1}=a_n+(\sqrt2-1) &\implies \\[2ex] -0.01+(\sqrt2-1)<a_{n+1}-L<0.01+(\sqrt2-1) &\implies \\[2ex] a_{n+1}-L>0.4 \end{align}$$

and in the second case

$$\begin{align} a_{n+1}=a_n+(\sqrt2-2) &\implies \\[2ex] -0.01+(\sqrt2-2)<a_{n+1}-L<0.01+(\sqrt2-2) &\implies \\[2ex] a_{n+1}-L<-0.5 \end{align}$$

So for any positive $\epsilon<0.01$, if for any supposed limit $L$ we have $|a_n-L|<\epsilon$, then we have $|a_{n+1}-L|>\epsilon$, so the limit does not exist.

Answer 2

I don't know if one can do that more formally, but this is how I would show it:

Denote the sequence by

$\{x_n\}_{n\in\mathbb{N}},\,x_n = n\sqrt 2-\lfloor n \sqrt{2}\rfloor$.

You're interested in the limit of this sequence.

Note: For $a\in\mathbb{R}$ you have that $a-\lfloor a\rfloor$ is the decimal-part of $a$. This decimal-part is always in $[0,1)$. Especially does that mean, that $\{x_n\}\subset [0,1)$.

Take some point $x\in [0,1]$ (if your series converges, it converges to some point in [0,1] ).

We need to show, that $\forall\varepsilon>0\exists n_0\in\mathbb{N}$ s.t. $|x_n-x|<\varepsilon,\,\forall n\geq n_0$ (or that this doesn't hold)

W.l.o.g we consider $\varepsilon < \frac{1}{2}$. If it fails to find $n_0$ for this type of epsilon, that's enough to prove that the series does not converge.

So, by restricting the choice of $\varepsilon$, either $x-\varepsilon$ or $x+\varepsilon$ is also in $[0,1)$.

W.l.o.g we have $\tilde{x} = x+\varepsilon\in[0,1)$.

$\sqrt{2}$ is irrational. This means that for our $x$, we can find an $n_0^x\in\mathbb{N}$ such that

$|n\sqrt{2}-\lfloor n\sqrt{2}\rfloor - x|<\frac{\varepsilon}{2}$.

However, we can also find an $n_0^\tilde{x}\in\mathbb{N}$ such that

$|n\sqrt{2}-\lfloor n\sqrt{2}\rfloor -\tilde{x}|<\frac{\varepsilon}{2}$.

HERE COMES THE VERY IMPORTANT PART: Again, as $\sqrt{2}$ is irrational, the decimal part of $n\sqrt{2}$ "bounces" in the interval (0,1). Especially, you will NOT find $m\in\mathbb{N}$ s.t. $m\sqrt{2}-\lfloor m\sqrt{2}\rfloor \in \{ 0,1\}$. Using this, we can always chose $n_0^\tilde{x} > n_0$. But the same argument also holds in the other direction, so we can always chose $n_0^x > n_0^\tilde{x}$.

You see:

$\forall \varepsilon<\frac{1}{2},\forall x\in[0,1] \not\exists n_0\in\mathbb{N}:\,|x_n-x|<\varepsilon,\,\forall n\geq n_0$.

This suffices to see, that $x_n$ does not converge.

Answer 3

According to Weyl's Criterion, the sequence $x_n$ is equidistributed if and only if

$$\lim_{N \to \infty} \frac{1}{N}\sum^{}_{n<N}e^{2\pi imx_n} = 0 \quad \mbox{for all} \quad m \in \mathbb{N}$$

A consequence of this result is that the sequence $nx-\lfloor{nx}\rfloor$ is dense for any irrational $x$.

That means a limit cannot exist since $\sqrt{2}$ is irrational.