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

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For the concrete example: $$\inf_{n \geq k} \, 2 - \frac{1}{n} = 2 - \frac{1}{k}$$ 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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### cnmesr

Updated on August 01, 2020

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