finding the limit superior of a sequence involving fractional part

Question

I'm stuck with this exercise that's asking to prove that:

$$\limsup_n\{\sqrt n\}=1$$

I've managed to prove that the limit inferior is $0$ but I've failed to prove this part.

Answer 1

Since $0\leq \{x\}<1$ for any real $x$, it follows that $$\limsup_n\,\{\sqrt n\}\leq 1$$ In order to show the equality, it suffices to find a sequence of positive integers $(n_k)_k$ such that $n_k\to +\infty$ and $\{\sqrt n_k\}\to 1$.

Take $n_k:=k^2-1$ for $k\in\mathbb{N}^+$. Then $k-1\leq \sqrt{k^2-1}< k$ implies that, as $k\to\infty$, \begin{align*} \{\sqrt{n_k}\}&=\sqrt{k^2-1}-(k-1)=k\sqrt{1-\frac{1}{k^2}}-k+1\\ &=k\left(1-\frac{1}{2k^2}+o(1/k^2)\right)-k+1=1+o(1)\to 1 \end{align*} where we used the fact that $\sqrt{1+x}=1+\frac{x}{2}+o(x)$ as $x\to 0$.

Answer 2

Hint: $$\sqrt{k²+2k}-k=\frac{2k}{\sqrt{k^2+2k}+k}$$