Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$

Question

The question is to find $\sup$ and $\inf$ of $B=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$ where $\lfloor x \rfloor$ is defined as the largest integer that is smaller than $x$.

That $\inf B=0$ is trivial, I'm having troubles with the other side though. I think intuitively that $\sup B=1$ and random calculations seems to confirm it since I can get as many 9's after the dot as I want by calculating $\sqrt{10^{2n}-1}$ for increasingly large values of $n$, but apart from that I have no idea how to get specific values or even know if a value is sufficient for a given $\epsilon$.

Any hints on where to go with such proofs?

Answer 1

Hint If $n=k^2+a$ where $0 \leq a <2k+1$ then $$ \sqrt{n}-\lfloor \sqrt{n} \rfloor=\sqrt{n}-k=\frac{n-k^2}{\sqrt{n}+k}=\frac{a}{\sqrt{n}+k}\geq \frac{a}{(k+1)+k}= \frac{a}{2k+1}$$

Now it is easy to chose $k$ and $0 \leq a \leq 2k$ such that $\frac{a}{2k+1}$ becomes arbitrarily close to $1$.

Answer 2

You should try to compute this limit: $$\lim_{n\to\infty}\left(\sqrt{(n+1)^2-1}-n\right)$$

If you don't know about limits, just write

$$\begin{align} \left(\sqrt{(n+1)^2-1}-n\right)&\cdot\frac{\sqrt{(n+1)^2-1}+n}{\sqrt{(n+1)^2-1}+n}\\ &=\frac{(n+1)^2-1-n^2}{\sqrt{(n+1)^2-1}+n}\\ &=\frac{2n}{\sqrt{n^2+2n}+n}\\ &=\frac2{\sqrt{1+\frac2n}+1} \end{align}$$

Note that the term $2/n$ becomes arbitrarily small as $n$ increases (informally speaking).