This is from an exercise given by the teacher.
Find the infimum and supremum of:
$$ A = \{n\in\mathbb{N} : \frac{4}{\pi}\arctan\left(1-n^{2}\right)+\frac{\left(1-2n^{2}\right)}{1+2n^{2}}\} $$
I have divided this into two subsequences using the fact that the limit of (a + b) is equal to limit of a + limit of b. I ended up with:
$$ \lim_{n \to \infty}\bigg(\frac{4}{\pi}\arctan\left(1-n^{2}\right)+\frac{\left(1-2n^{2}\right)}{1+2n^{2}}\bigg) =-3 $$
And I know that if the sequence is monotonically increasing (decreasing) then the lower (upper) limit is equal to $a_1$.
So:
$$ a_1 = -\frac{1}{3} $$
But I don't know if this this sequence is decreasing or increasing. I have tried to approach it from the angle that a sum of two decreasing (increasing) sequences is decreasing (increasing). So I've decided that
$$ \frac{4}{\pi}\arctan\left(1-n^{2}\right) $$
is decreasing as n goes to infinity and that leaves me with only the second "part" of the sum.
$$\frac{1-2n^{2}}{1+2n^{2}}$$
And now I have to check if
$$ \frac{1-2(n + 1)^{2}}{1+2(n+1)^{2}} - \frac{1-2n^{2}}{1+2n^{2}} < 0 $$
Wolfram said that it's less than $0$ if $n > -\frac{1}{2}$ and because n is natural then it's $n \ge 1$. So both of the "parts" of the sum are monotonically decreasing, so the upper bound of A is $-\frac{1}{3}$ and the lower bound is $-3$.
My question right now is, did I have to think about its monotonicity or did I only have to determine $a_1$ and say that it's bigger than $-3 $ and from that just say that the smaller one is the lower bound and the bigger one is the upper bound?