Supremum of rounded function Can you please help me to find and prove $\sup\{\sqrt{n}- \left\lfloor \sqrt{n}\right\rfloor : n\in N\}$?
I assume that it is something like $1$ or some other constant, but not quiet sure.

To prove that $$\sup \left\{ \sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor  : n \in \Bbb{N}  \right\} = 1$$
(note that I use the more modern symbol $\lfloor x \rfloor$ in place of $[x]$ to mean the highest integer not exceeding $x$)
Step 1: For $n \in \Bbb{N}$  or indeed any real value of $n$, $\sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor <1$.  Because if $\sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor  = 1+\alpha$ with $\alpha > 0$, then $\left\lfloor \sqrt{n} \right \rfloor +1  = \sqrt{n}-\alpha < \sqrt{n}$ so $\left\lfloor \sqrt{n} \right \rfloor$ is not the greatest integer not exceeding $\sqrt{n}$ and this contradicts the definition of  $\lfloor x \rfloor$. 
Step 2: For any $\epsilon >0$, there exists some $n \in \Bbb{N}$ such that 
$$
\sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor > 1-\epsilon
$$
Proof: Choose any $k > \frac1\epsilon + 1$ and consider $n = k^2 -1$. Then
$\left\lfloor \sqrt{n} \right \rfloor = k-1$ and $\sqrt{n} = (k-1) + x$ for some $0<x<1$. Now of any $x : |x|\leq 1$, 
$$
1+\frac{x}2-\frac{x^2}8 = \sqrt{1+x-\frac{x^3}{8}+\frac{x^4}{64}} < \sqrt{1+x}
$$
In particular, choose $x = \frac{2}{k-1}$ so that 
$$
\sqrt{1+\frac2{k-1}}> 1+\frac1{k-1}-\frac1{2(k-1)^2} \\
(k-1)\sqrt{1+\frac2{k-1}}> (k-1)\left(1+\frac1{k-1}-\frac1{2(k-1)^2} \right)\\
\sqrt{k^2-2k+1+2(k-1)}> k-1+1-\frac1{2(k-1)} \\
\sqrt{k^2-1)}> k-\frac1{2(k-1)} > k-\frac1{2/\epsilon} = k-\frac{\epsilon}{2} > k-\epsilon
$$
So for our arbitrary $\epsilon$ we have demonstrated an $n$ such that 
$$\left\{ \sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor \right\}> 1-\epsilon$$
The combination of steps 1 and 2 establishes that $1$ is the desired suprenum.
