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Calculate supremum, infimum, lim sup and lim inf of $a_{n}=2-\frac{1}{n}$ for $n \in \mathbb{N} $

$\sup = 2-\left(\lim_{n\rightarrow\infty}\frac{1}{n}\right)=2-0 = 2$

$\inf = 2-\frac{1}{1}=2-1=1$

$\limsup_{n\rightarrow\infty}\left(2-\frac{1}{n}\right)=2-0=2$

$\liminf_{n\rightarrow\infty}$ I'm not sure about that but I would say it doesn't exist, it will be 2, same as $\limsup$.

Is it alright?

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    $\begingroup$ You are correct. This is general, in the sense that the limit exists iff lim sup and lim inf are equal (lim sup and lim inf allways exist, but they may not be finite). $\endgroup$ – VictorZurkowski Jul 31 '16 at 10:19
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VictorZurkowski already answered correctly in his comment.

For the concrete example: \begin{equation} \inf_{n \geq k} \, 2 - \frac{1}{n} = 2 - \frac{1}{k} \end{equation} So taking $\lim_{k \rightarrow \infty} 2 - \frac{1}{k} = 2$.

Thus, $\lim \inf_{n \rightarrow \infty} 2 - \frac{1}{n} = 2$.

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  • $\begingroup$ Does that mean lim inf = 2 too? Or I better say doesn't exist? $\endgroup$ – cnmesr Jul 31 '16 at 10:41
  • $\begingroup$ lim inf = 2, as VictorZurkowski already mentioned. $\endgroup$ – ChrisT Jul 31 '16 at 11:23

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