Prove $\sup\{(-1)^n+\frac{2}{n}\leq a: n\in\mathbb{N},a\in(-1,3)\}$ is inside the set which its supremum of

Question

Prove $\sup\{(-1)^n+\frac{2}{n}\leq a: n\in\mathbb{N},a\in(-1,3)\}$ is inside the set which its supremum of

What I want to achieve is $\exists n\in\mathbb{N},(-1)^n+\frac{2}{n}\leq a<(-1)^n+\frac{2}{n-1}$. That's why I created the set as seen in the title.

As you can see, I am having a problem in proving the supremum belongs to its set and I need this fact to justify the inequality I stated above.

So is there any elegant way to prove this? I have an approach.

1. The set of that supremum has a bijection with set of natural numbers
2. I can ensure the set of natural numbers involved are bound below
3. Hence, by axiom of completeness, there exists infimum

This is just a brief idea but I don't know how to write it out formally. I would also welcome other ideas, especially the general idea of proving supremum belongs to its set.

Answer 1

Using limits we can show that when $n \rightarrow \infty $ the sequence becomes $-1$ or $1$,

By using a ratio of consecutive terms you can show that the sequence decreases with $n$ Then we have Supremum $ = 1 \in (-1,3)$ This is not the case with the infimum as $-1 \notin (-1,3) $