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Given $a_{n}=(-1)^{n}n$, I am trying to compute

a) $\lim\limits{\inf(a_{n})}$

b) $\lim\limits{\sup(a_{n})}$

c) $\inf\{a_{n}\}$

d) ${\sup(a_{n})}$

My work:

a) $\lim\limits{\inf(a_{n})}=-\infty$

b) $\lim\limits{\sup(a_{n})}=+\infty$

c) $\inf\{a_{n}\}=-\infty$

d) ${\sup(a_{n})}=+\infty$

I find it hard work and results formally prove that these results only the intuition

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Your answers look right. Justifying your infimum and supremum should not be hard. What are your definitions of $\limsup$ and $\liminf$? – user21436 May 1 '12 at 5:24
Start with the $\sup$ and $\inf$ first. For example, $\sup \{ a_n \}$ is an upper bound for $a_n$. So it must be $\geq$ all $a_n$. So choose specific $n$ that show why the answer must by $+ \infty$. – copper.hat May 1 '12 at 5:25
Many of the ideas that I suggested in this answer to your earlier question can be applied to this problem as well. – Brian M. Scott May 1 '12 at 6:44
up vote 0 down vote accepted

I'll help you justify why the supremum is $+\infty$ and leave the part about infimum to you.

Let $A \subseteq \Bbb R$. We say that the $\sup A=+\infty$ if $A$ is not bounded above.

Claim: $\{a_n\}_{n \geqslant 1}$ is not bounded above.

Let $b \in \Bbb R^+$. Consider $\lfloor b \rfloor +i$, where $i=1$ if $\lfloor b\rfloor$ is odd and $i=2$ if $\lfloor b\rfloor$ is even. Look at the definition of $\lfloor b \rfloor$: $$b-1\lt\lfloor b \rfloor \le b$$

So, $b \lt\lfloor b \rfloor +1 \leq \lfloor b \rfloor +i$ for our choices of $i$. Can you show that $a_{\lfloor b \rfloor+i}= \lfloor b \rfloor +i$?

So, you have found a $n$ such that $b \lt a_n$. So, is $b$ an upper bound? What do you conclude?

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@DanieldelCarmen I can help you about the $\liminf$ and $\limsup$ if only I know your deefinitions. So, please add them in. Regards, – user21436 May 1 '12 at 5:50

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