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Given $a_{n}=\frac{(-1)^{n}n+1}{n}$, compute $$\lim\limits{\inf(a_{n})}$$ $$\lim\limits{\sup(a_{n})}$$ $$\inf\{a_{n}\}$$ $${\sup(a_{n})}$$

My attempt: I tried taking different values ​​for the sequence and reached the following results: $$\lim\limits{\inf(a_{n})}=1$$ $$\lim\limits{\sup(a_{n})}=-1$$ $$\inf\{a_{n}\}=3/2$$ $${\sup(a_{n})}=-1$$

The teacher told me to do it more formally, anyone can help me please?

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Remember that $\liminf a_n \leq \limsup a_n$. –  Siminore Apr 30 '12 at 15:21
    
You have accidentally swapped $\inf$ and $\sup$ in all of the above, but otherwise these would be the correct answers. You just need to prove it, that is provide a rigorous justification. –  Eric Naslund Apr 30 '12 at 15:24
    
It's all correct, just remember that sup is the large one and inf the small. –  Ekuurh Apr 30 '12 at 15:24

2 Answers 2

up vote 4 down vote accepted

Note that you've interchanged '(lim)inf' and '(lim)sup' throughout your question.

It's helpful to note that

$$a_n=\begin{cases} 1+\frac1n,&\text{if }n\text{ is even}\\\\ -1+\frac1n,&\text{if }n\text{ is odd}\;. \end{cases}$$

Let's look first at $\sup_n a_n$. When $n$ is odd, $a_n\le 0$, and when $n$ is even, $a_n>0$, so in order to find $\sup_n a_n$ we need only look at the even-numbered terms: all of them are larger than any of the odd-numbered terms. But the subsequence $\langle a_{2n}:n\in\Bbb Z^+\rangle$ is clearly decreasing, so for every $n\in\Bbb Z^+$ we have $a_n\le a_2=\frac32$. Thus, $\sup_n a_n=\max_n a_n=a_2=\frac32$.

You can make a similar argument for $\inf_n a_n$.

Now let's look at $\limsup_n a_n$. By definition $$\limsup_n a_n=\lim_{n\to\infty}\sup_{k\ge n}a_k\;,\tag{1}$$ so for each $n\in\Bbb Z^+$ we need to see what $\sup\{a_k:k\ge n\}$ is. For this we can reason just as I did above to find that

$$\sup_{k\ge n}a_k=\begin{cases} a_n,&\text{if }n\text{ is even}\\ a_{n+1},&\text{if }n\text{ is odd}\;,\tag{2} \end{cases}$$

but I'll leave the details to you. Once you've shown $(2)$, you just have to evaluate $(1)$, and you should find that it's simply $$\lim_{n\to\infty}a_{2n}=\lim_{n\to\infty}\left(1+\frac1n\right)=1\;.$$ (I doubt that you'll be asked for a formal proof that this limit really is $1$.)

The arguments that you need for $\liminf_n a_n$ are very similar.

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To make it more formal just use the definitions that you have for lim inf, lim sup, inf, and sup. Show that your answers meet those definitions.

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