Limit of sequence question with confusing notation

[tomtom690]

Homework Statement

Hello, I don't entirely understand the question here. It is the limit of a sequence, however it is defined on what appears to be an interval, and is different if n is odd or even. Excuse the messy notation here, I don't know how to use LaTeX.

Let the sequence {A_n : n>=1} subset of P(Omega) be defined by:
A_n={( (n^2-1)/3n^3+1 , 4n^2/n^2+1 ] n even
{( 5n^2/(2n^2-3) , (7n^2+5)/3n^2 ] n odd

Study if lim A_n, n-->infinity exists.
In words, the n even case is "the interval open on the left and closed on the right from n squared minus one over three n cubed plus one, to four n squared over n squared plus one."
Its the interval bit which doesn't make any sense to me!

Homework Equations

This is part of an advanced probability course (hence the P(Omega)) if this means anything.

The Attempt at a Solution

Well, first I tried evaluating limits as n goes to infinity on all four expressions, hoping that I would get numbers which would be the same in the odd and even case. I used l'hopitals rule in all cases, and for the evens got an interval of (1/3, 4] and the odds of (5/2, 7/3] which is rediculous, for starters!
Thanks in advance

 

[tiny-tim]

welcome to pf!

hi tomtom690! welcome to pf!

(have an infinity: ∞ and an omega: Ω and try using the X² and X₂ icons just above the Reply box )

Let the sequence {A_n : n>=1} subset of P(Omega) be defined by:
A_n={( (n^2-1)/3n^3+1 , 4n^2/n^2+1 ] n even
{( 5n^2/(2n^2-3) , (7n^2+5)/3n^2 ] n odd

Study if lim A_n, n-->infinity exists.
…
Well, first I tried evaluating limits as n goes to infinity on all four expressions, hoping that I would get numbers which would be the same in the odd and even case. I used l'hopitals rule in all cases, and for the evens got an interval of (1/3, 4] and the odds of (5/2, 7/3] which is rediculous, for starters!

well, half the sequence converges to one point in P(Ω), and the other half converges to a different point in P(Ω), so … ?