# Thread: Find the superior and inferior limit??

1. ## Find the superior and inferior limit??

Hi guys:
the Qs is 'Find the superior and inferior limit of the sequence"(-1)^n(1+n^-1)". Even I know the answer for the Qs is 1 and -1, but what should be the currect process for this Qs, please give some detail.thanks!! <img smilieid="84" class="inlineimg" src="http://www.mathhelpforum.com/math-help/images/smilies/Skype/emoticon-0139-bow.gif" border="0">

2. Originally Posted by wsun
Hi guys:
the Qs is 'Find the superior and inferior limit of the sequence"(-1)^n(1+n^-1)". Even I know the answer for the Qs is 1 and -1, but what should be the currect process for this Qs, please give some detail.thanks!! <img smilieid="84" class="inlineimg" src="http://www.mathhelpforum.com/math-help/images/smilies/Skype/emoticon-0139-bow.gif" border="0">
I give you a theorem to judge the superior limit of a sequence.
Let $\displaystyle E=\{a|\exists subsequence\{b_n\}, \lim b_n=a\}$,and $\displaystyle +\infty,-\infty$may also appear in $\displaystyle E$.
Then $\displaystyle b$is the superior limit of $\displaystyle \{b_n\}$iff
1: $\displaystyle b\in E$
2: $\displaystyle \forall x>b, \exists N, \forall n>N, a_n<x$
For the inferior limit, it is similiar.
Now you can easily identify whether the answer is 1 or -1..

3. Your sequence is given by $\displaystyle a_n= (-1)^n(1+ \frac{1}{n})$. Calculate a few values: $\displaystyle a_1= -2$, $\displaystyle a_2= \frac{3}{2}$, $\displaystyle a_3= -\frac{4}{3}$, $\displaystyle a_4= \frac{5}{4}$.

In fact it should be clear that $\displaystyle a_n= (-1)^n(1+ \frac{1}{n})= (-1)^n\frac{n+1}{n}$. If n is odd that is $\displaystyle a_n= -\frac{n+1}{n}$ which converges to -1 and if n is even that is $\displaystyle a_n= \frac{n+1}{n}$ which converges to 1. Since, if a sequence converges, every convergent subsequence must converge to the same thing, it is clear that every convergent subsequence converges to either 1 or -1.

Now, "lim sup" is defined as the supremum (least upper bound) of the set of subsequential limits while the "lim inf" is the infimum. Here, the set of subsequential limits is {-1, 1} which has minimum -1 and maximum, 1 and, since it has a minimum and maximum, those are the "inf" and "sup" and so the lim inf and lim sup.

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# how to find limit supremum

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