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$\begingroup$

$a_{n}=(-1)^n$ if n is even,$ a_{n}=\sqrt{n}$ if n is odd

infimum is 1

supremum is ∞ (if we allow ∞ as supremum. If we don't allow then we say supremum does not exist) because √n increases to infinity

for any m, the set an where n>=m has infimum 1 and supremum ∞, so the limit inferior is 1 and limit superior is ∞

Is it right to my procedure?

Is there a way to make the result more formal?

1 Answers 1

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Yes, $\liminf_{n\to\infty} a_n = 1.$ The $\liminf$ is the value of the smallest limit point of the sequence. You have $\limsup_{n\to\infty} a_n = +\infty.$ since a subsequence of the $a_n$ increases without bound.

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    I first saw this result in the book of Saxena and Shah.2012-05-28