# Thread: lim sup, lim inf and cluster points.

1. ## lim sup, lim inf and cluster points.

Hi

So, lets say i have a sequence a(n). A sequence can have a lim superior (greatest cluster point) or lim inferior (lowest cluster point), only if the sequence is bounded, right?

so, given a(n):= (1 + (-1)^n nē) / (2 + 3n + nē)
with n >> infinity, lim a(n) diverges between -infinity and infinity therefor it isn't bounded and has no lim superior or lim inferior. correct?

and a more tricky sequence definied by:
a(3n-2):= 3 + 1/n
a(3n-1):= 2/n
a(3n):= - 1/nē
my (correct/incorrect) answer:
the sequence a(n) has two cluster points, a limit inferior of the value 0 and a limit superior of the value 3.

is everything i mentioned above correct?

2. ## Re: lim sup, lim inf and cluster points.

Originally Posted by nappysnake
sequence a(n). A sequence can have a lim superior (greatest cluster point) or lim inferior (lowest cluster point), only if the sequence is bounded, right?

so, given a(n):= (1 + (-1)^n nē) / (2 + 3n + nē)
with n >> infinity, lim a(n) diverges between -infinity and infinity therefor it isn't bounded and has no lim superior or lim inferior. correct?
Surely $\displaystyle \frac{{1 + ( - 1)^n n^2 }}{{2 + 3n + n^2 }} = \frac{{\tfrac{1}{{n^2 }} + ( - 1)^n }}{{\tfrac{2}{{n^2 }} + \tfrac{3}{n} + 1}}~??$

If I have read the question correctly, you will want to rethink the first part.

3. ## Re: lim sup, lim inf and cluster points.

Oops, the sequence diverges between -1 and 1 therefor possesses a lim superior and a lim inferior..my bad. How about the second sequence, everything alright with it?