On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$

Question

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 $3= \sup(A)$.

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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