Calculate limit with floor function 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.
Let $a_n=n\sqrt 2-\lfloor n\sqrt2 \rfloor$. Note that $a_n&gt;0$ 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 $N\in\mathbb{N}$),

$$0.01&lt;a_n-L&lt;0.01$$

Then

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

In the first case

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

and in the second case

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

So for any positive $\epsilon&lt;0.01$, if for any supposed limit $L$ we have $|a_n-L|&lt;\epsilon$, then we have $|a_{n+1}-L|&gt;\epsilon$, so the limit does not exist.
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 $\forall\varepsilon&gt;0\exists n_0\in\mathbb{N}$ s.t. $|x_n-x|&lt;\varepsilon,\,\forall n\geq n_0$ (or that this doesn't hold)

W.l.o.g we consider $\varepsilon &lt; \frac{1}{2}$. 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

$|n\sqrt{2}-\lfloor n\sqrt{2}\rfloor - x|&lt;\frac{\varepsilon}{2}$.

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

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

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 $n_0^\tilde{x} &gt; n_0$. But the same argument also holds in the other direction, so we can always chose $n_0^x &gt; n_0^\tilde{x}$.

You see:

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

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