$a_n= \sqrt{n^2-1}-n$ superior and inferior limits

I'm given the succession $$a_n= \sqrt{n^2-1}-n$$ and I should find the superior limit $$\limsup_{n \to \infty}a_n$$ and the inferior limit $$\liminf_{n \to \infty}a_n$$

This succession is defined $$\forall n \ge 1$$ , $$n \in N$$

if $$n=1, a_1=-1$$;

if $$n=2, a_2=\sqrt{3}-2$$;

if $$n=3, a_3=\sqrt{8}-3$$;

if $$n=4, a_3=\sqrt{15}-4$$

....

$$\sqrt{n^2-1}< n \rightarrow \sqrt{n^2-1}- n <0 \rightarrow a_n<0, \forall n \ge1$$

$$\lim_{n \to \infty}a_n = 0$$

and considering the real function , its derivative is strictily positive.

In my opinion it should be $$\limsup_{n \to \infty}a_n=0$$

$$\liminf_{n \to \infty}a_n=-1$$

Is it right?

• Write it as $\sqrt{n^2-1}-n=\left(\sqrt{n^2-1}-n\right)\frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}=\frac{-1}{\sqrt{n^2-1}+n}$ Dec 27, 2019 at 15:26
• In my opinion, they're both $0$ Dec 27, 2019 at 15:28
• You can evaluate the regular limit of the sequence which would imply the limit superior and inferior are equal to it. Dec 27, 2019 at 15:29
• @bjorn93 so if the succession has a limit,is this limit the superior and inferior limit?
– Anne
Dec 27, 2019 at 15:49
• The limit of a sequence exists if and only if both its limit superior and limit inferior exist and they are equal Dec 27, 2019 at 15:50

Since$$\lim_{n\to\infty}\sqrt{n^2+1}-n=\lim_{n\to\infty}\frac1{\sqrt{n^2+1}+n}=0,$$you have$$\limsup_{n\to\infty}\sqrt{n^2+1}-n=\liminf_{n\to\infty}\sqrt{n^2+1}-n=0.$$