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$a_{n}=(-1)^n$ 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?

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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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  • $\begingroup$ Hello, Do you hear there would be a more formal way of doing it? $\endgroup$ – Daniela del Carmen May 28 '12 at 0:49
  • $\begingroup$ $\liminf_{n\to\infty} a_n = \sup_{ N} \inf_{n \ge N} a_n$. Since $a_n = 1$ infinitely often, $\inf_{n\ge N} a_n = 1$ for all $n$. You can deal similarly to see how the $\limsup$ works. BTW, the result I cite is widely known. If you are studying $\limsup$ and $\limsup$, see if you can prove it. $\endgroup$ – ncmathsadist May 28 '12 at 0:57
  • $\begingroup$ I can't believe I never realized that the lim inf is the least cluster point of a sequence! Thanks for your answer! $\endgroup$ – Cameron Buie May 28 '12 at 2:10
  • $\begingroup$ I first saw this result in the book of Saxena and Shah. $\endgroup$ – ncmathsadist May 28 '12 at 2:11

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