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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?

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

1 Answer 1

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

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