Finding the supremum and infimum of a set of natural numbers 
Let $A = \left\{ 2(-1)^{n+1} + (-1)^{ \frac{n(n+1)}{2} } \left( 2 +
 \dfrac{3}{n} \right) : n \in \mathbb{N} \right\}.$ Our goal is find $\sup A$ and $\inf
 A$.

Attempt
At first glance, it looks like a very formidable set, and I can't see any obvious way but to separate in cases whether $n$ is odd or even.
If $n=2k$, then $n+1$ is off and $\dfrac{n(n+1)}{2} = k(2k+1)$ and thus our set takes the form
$$ A = \{ -2 + (-1)^k (2 + 3/2k) : k \in \mathbb{N} \} $$
We observe that for large values of $k$, the value $3/(2k)$ is negligible and so we have $-2 + (-1)^k 2$. In other words, we can have either $-4 $ or $0$ and so we claim that
$$ \sup A = 0 \; \; \text{and} \; \; \inf A = -4 $$
I am getting stuck on trying to actually prove these claims rigorously. Can I get some advice in how to do so?
Update:
Maybe itd be easy if we write $A_1 = \{ 2 (-1)^{n+1} \}$ and $A_2 = \left\{ (-1)^{ \frac{n(n+1)}{2} } \left( 2 +
 \dfrac{3}{n} \right) \right\} $
and use $\sup(A_1 + A_2) = \sup A_1 + \sup A_2 $ may help

Let 
$$F(n) = 2(-1)^{n+1} + (-1)^\frac{n(n+1)}{2}\left(2+\frac{3}{n}\right)$$
For $n \ge 3$ we have $| F(n)| \le 5$, so $-5 \le F(n) \le 5$ when $n \ge 3$.
Since $F(1) = -3$, $F(2)=-5.5$ and $F(3)=5$, we have $\sup A = 5$ and $\inf A = -5.5$.
