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

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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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Updated on August 01, 2020

Comments

  • cnmesr
    cnmesr over 3 years

    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?

    • VictorZurkowski
      VictorZurkowski over 7 years
      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).
  • cnmesr
    cnmesr over 7 years
    Does that mean lim inf = 2 too? Or I better say doesn't exist?
  • ChrisT
    ChrisT over 7 years
    lim inf = 2, as VictorZurkowski already mentioned.