<s>[INST] 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 $3= \sup(A)$. 
 [/INST] 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.
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