Find the inf and supremum

[Mathyes]

Some one please help explain the second part of this problem

Problem: Find the inf and sup for the subset of \(\displaystyle \Re \)

Given: \(\displaystyle \{ {( - 1)^n}\frac{{n - 6}}{{n + 3}}|n \in \aleph \} \)

Solution: When n = 1 then plug into the formula, I will have 5/4 which is my least upper bound (sup). But what about when n approach infinity, I cannot calculate it to have the (greatest lower bound = -1 as told by the solution sheet). I am not sure if I consider n approaches infinity is right. I just figure the natural number range from 1, 2, 3, ...,

Is this true for me to find the inf as follow ?

Take the limit of the given as: limit of \(\displaystyle \{ {( - 1)^n}\frac{{n - 6}}{{n + 3}}\} \) as \(\displaystyle n \to \infty \)

I know that the the limit of \(\displaystyle \frac{{n - 6}}{{n + 3}} = 1\) as \(\displaystyle n \to \infty \). Then what about the limit of \(\displaystyle {( - 1)^n}\) as \(\displaystyle n \to \infty \), I just can't take the limit of that. How can I determine the sign of 1 if it is positive or negative?

 

[pka]

Problem: Find the inf and sup for the subset of \(\displaystyle \Re \)
Given: \(\displaystyle \{ {( - 1)^n}\frac{{n - 6}}{{n + 3}}|n \in \aleph \} \)
Solution: When n = 1 then plug into the formula, I will have 5/4 which is my least upper bound (sup). But what about when n approach infinity, I cannot calculate it to have the (greatest lower bound = -1 as told by the solution sheet). I am not sure if I consider n approaches infinity is right. I just figure the natural number range from 1, 2, 3, ...,

When \(\displaystyle n=(2k-1)\) is an odd number, the corresponding term is \(\displaystyle \dfrac{-(2k-7)}{2k+2}\).

As \(\displaystyle n\to\infty\) it follows that \(\displaystyle k\to\infty\) or \(\displaystyle \dfrac{-(2k-7)}{2k+2}\to -1\).

 

[Mathyes]

Thanks for the help, I got it.