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Prove that if $\{a_n\}$ is a sequence, then $\limsup a_n$ and $\liminf a_n$ are subsequential limits of $\{a_n\}$.

I don't know the case where $\limsup a_n = \infty$.

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  • $\begingroup$ Same as for finite limsup. $\endgroup$
    – Berci
    Oct 20, 2012 at 19:22
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    $\begingroup$ Please don't write the whole question in the title. This is what the body is for. $\endgroup$
    – user2468
    Oct 20, 2012 at 19:46
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    $\begingroup$ We can actually define the limit superior and limit inferior of a sequence as the greatest and least limit point of the sequence, meaning they are subsequential limits of the sequence. With an alternate definition of them, you can prove they are indeed limit points and the if $x$ is any other limit point, it is between them. $\endgroup$
    – Pedro
    Jan 11, 2013 at 23:14

2 Answers 2

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$$\limsup a_n = \inf_{k}\big( \sup_{n\ge k} a_n\big)$$ This is $+\infty$ iff $\forall k:\ \sup_{n\ge k} a_n=+\infty$. Then for each given previous index $k$ and $z\in\Bbb N$, there is a next index $n\ge{k+1}$ such that $a_n>z$. This selected subsequence is thus bigger than the sequence $1,2,3,4,5,..$, hence its limit is $+\infty$.

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For the case you asking for let's take this sequence: $$(-1)^{n}n$$ which is: $$\{-1,2,-3,4,-5 \cdots\}$$ for this sequence is $\liminf = - \infty$ and $\limsup = \infty$. This are limits of subsequences of odd and even terms respectivly.

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