On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$. 
Let $\epsilon >0$. We wish to prove that $3-\epsilon$ is not an upper bound for $A$. 
For $N = 2n_0$ where $\displaystyle n_0 >  \frac{1}{\sqrt{2\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} =  3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$. 
EDIT: 
For $N = 2n_0$ where $\displaystyle n_0 >  \frac{1}{\sqrt{\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} =  3-\frac{1}{N^2+1} \ge 3-\frac{1}{N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$. 
 A: Looks good to me. EDIT - though as Martiny pointed out in the comments, it's not. You wrote $3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2}$ when this is not true.

You can also do this by taking every other term directly: noting that $$\lim_{n \to \infty} \left[3(-1)^{2n} - \frac{1}{(2n)^2+1} \right] = \lim_{n \to \infty} 3  - \lim_{n \to \infty} \frac{1}{(2n)^2+1} = 3 - 0 = 3$$
so the sequence must get arbitrarily close to $3$ through its even terms anyway.
A: It is fine. Only a technical detail.
$$3-\frac1{2N^2}=3-\frac1{4n_0^2}>3-\frac{2\epsilon}4$$
This is certainly greater than $3-\epsilon$, so your proof is correct, but it would be more elegant if you simply say that $N>1/\sqrt\epsilon$ and even.
A: This sequence        $$ A = \left\{3(-1)^n-\frac{1}{n^2+1}: n  \in \mathbb{N}\right\}$$ 
has a subsequence  $$  \left\{3-\frac{1}{4n^2+1}: n \in \mathbb{N}\right\}$$  which is monotonically increases to $x=3$
Since x=3 is an upper bound for the set $$ A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$$
We have $$ sup (A) = 3
