Sup and inf of $A=\{a_n:=\frac{(n^2+(-1)^n(n-1)^2}{3n^{(1+(-1)^n)}+2n},\ n\in\mathbb{N}\setminus\{0\}\}$ I have to find the infimum and the supremum (if they exist) of A:
$$A=\{a_n:=\frac{(n^2+(-1)^n(n-1)^2}{3n^{(1+(-1)^n)}+2n},\ n\in\mathbb{N}\setminus\{0\}\}$$
I have worked as follows:
$1) a_1=\frac{1}{5},\, a_2=\frac{5}{16}, a_3=\frac{5}{9}, a_4=\frac{25}{16}$, where
$$a_1<a_3$$ and $$a_2<a_4$$
maybe the subsequence of the odd terms are both of them increasing? In this case
$$\exists \lim_{n\to\infty}a_{2n}=\frac{2}{3}=sup\{a_{2n}\}\,\,\,\,\text{and}\,\,\,\, \exists \lim_{n\to\infty}a_{2n+1}=1=sup\{a_{2n+1}\}$$
Then $$supA=max\{\frac{2}{3},1\}=1$$
$2)$ Since both of the subsequences are increasing then $$a_1<a_{2n+1}\, \forall n\geq 0$$
$$a_{2}<a_{2n}\, \forall n\geq 1$$
So
$$infA=min\{\frac{1}{5},\frac{5}{16} \}=\frac{1}{5}=minA$$
To do: can you tell me if my idea is right please? Anyway if it was, I should verify the sequence to be increasing!
 A: Your Idea is correct,
First let us calculate $a_{2n}$ and $a_{2n-1}$
$$ a_{2n} = \dfrac{8n^2-4n+1}{12n^2+2n} = \dfrac{2}{3} - \dfrac{2\left(\dfrac{n}{3}-\dfrac{1}{8}\right)}{3\left(n^2+\dfrac{n}{6}\right)} $$
And $$ a_{2n-1} = 1 - \dfrac{4}{4n+1} $$
Now it is very easy to verify that $a_{2n-1}$ is an increasing sequence, so infimum along the odd numbers is $1/5$ as you calculated, also the sup is $1$ which is also very easy to observe.
Claim: $a_{2n}$ is increasing.
Proof:
let $f(x) = \dfrac{x/3 - 1/8}{x^2 + x/6}$,
Then $f'(x) = \dfrac{48x^2+52x-3}{(12x^2+2x)^2} > 0$ if $x \ge 1$
So this just validates your calculations
A: Here, the general approach I use when I have to work on sequenceces with terms like $(-1)^n$.
Let $b_k$ the sequence define by:
$$b_k = \frac{4k^2+(2k-1)^2}{3\cdot 4k^2+4k}$$
where I replaced $n$ with $2k$, noticing $(-1)^{2k}=1$. As you noticed $b_k$ is monotonically increasing: you can prove it by induction or noticing that $b_{k+1}-b_k>0 \;\; \forall k\in \mathbf{N}$. Also:
$$\lim_{k \to +\infty}b_k = \lim_{k\to+\infty}\frac{4k^2+(2k-1)^2}{3\cdot 4k^2+4k}\sim\lim_{k\to+\infty}\frac{8k^2}{12k^2}=\frac{2}{3}\implies \sup_k\{b_k\}=\frac{2}{3}$$
And:
$$b_1=a_2=\inf_k\{b_k\}=\min_k\{b_k\}=\frac{5}{16}$$
With the same method, let:
$$c_i=\frac{(2i+1)^2-4i^2}{4i+5}$$
with $n=2i+1$, noticing $(-1)^{2i+1}=-1$. Also, $c_i$ is monotonically increasing (you can show it with the same method of $b_k \; (*)$. Taking the limit:
$$\lim_{i\to +\infty}\frac{(2i+1)^2-4i^2}{4i+5}\sim\lim_{i\to +\infty}\frac{4i}{4i}=1 \implies \sup_i\{c_i\}=1$$
And:
$$c_0=a_1=\min_i\{c_i\}=\inf_i\{c_i\}=\frac{1}{5}$$
Notice, aslo, that $\max_i\{c_i\}$ and $\max_k\{b_k\}$ are not defined.
Now, comparing, we can conclude that:

*

*$a_n$ has global minimum: $$\min_n\{a_n\}=\inf_n\{a_n\}=\frac{1}{5}$$

*$a_n$ has superior extreme, but not global maximum:
$$\sup_n\{a_n\}=1\;\;\, \nexists \max_n\{a_n\}$$
Your calculations are correct.
For completness, I'll show $(*)$:
$$c_{i+1}-c_i=\frac{4\cdot(i+1)+1}{4\cdot(i+1)+5}-\frac{4i+1}{4i+5}=\frac{(4i+5)^2-(4i+1)\cdot(4i+9)}{(4i+9)\cdot(4i+5)}=\frac{16}{(4i+9)\cdot(4i+5)}>0\,\;\forall i\in \mathbf{N}$$
$c_i$ is monotonically increasing.
