I have to find the extrema of: $$a_n=(-1)^{n+1}\left(2-\frac{n+2}{2^n}\right)$$ I calculated $a_1= \frac{1}{2}$ and $\displaystyle \lim_{n} a_n=(-2,2)$. Then $$\left|\left(2-\frac{n+2}{2^n}\right)\right|<2$$
thus I think that $\sup=2$ and $\inf=-2$ but I am not sure how to demonstrate it.
My attempt:
I separate $a_n$ into $a_{2k}$ and $a_{2k-1}$, then:
$$a_{2k}<2 \Rightarrow 1\left(2-\frac{2k+2}{2^{2k}}\right)<2, \text{ true for } k>-1$$
$$a_{2k-1}>-2 \Rightarrow -1\left(2-\frac{2k+1}{2^{2k-1}}\right)>-2, \text{ true for } k>-\frac1{2}$$
So I know that $a_n$ is bounded.
I tried to demonstrate that $\sup=2$ and $\inf=-2$ by doing the following:
$$a_{2k}>2-\varepsilon \Rightarrow 1\left(2-\frac{2k+2}{2^{2k}}\right)>2-\varepsilon$$
$$a_{2k-1}<-2+\varepsilon \Rightarrow -1\left(2-\frac{2k+1}{2^{2k-1}}\right)<-2+\varepsilon,$$ but I can't solve those inequalities. Is this one the right approach?
EDIT
Another attempt: (Thanks to G Cab's answer)
$$-b_n<a_n<b_n$$
where $b_n=\left(2-\frac{n+2}{2^n}\right)$.
$-b_n$ is decreasing and the limit is $-2$, and as the sequence is decreasing, $-2=\inf(-b_n)$.
$b_n$ is increasing and the limit is $2$, and as the sequence is increasing, $2=\sup(b_n)$. Thus
$$-2<-b_n<a_n<b_n<2$$
and $\sup(a_n)=2$ and $\inf(a_n)=-2$
Note that $$ \eqalign{ & f(x) = 2 - {{x + 2} \over {2^{\,x} }}\quad \Rightarrow \quad y(0) = 0 \cr & f'(x) = {{\ln 2\left( {x + 2} \right) - 1} \over {2^{\,x} }} \cr & 0 < f'(x) = \quad \Rightarrow \quad {1 \over {\ln 2}} - 2 \approx - 0.557 < x \cr} $$ so $f(n)$ is increasing for $0 \le n$.
Thus the alternating sequence $(-1)^{n+1}f(n)$ is clearly bounded to stay between $\pm 2$.
I am not a theorist, but I think that a formal demonstration should develop along these lines
(maybe somebody more expert can help and put it in right terms)
Given that $$ \left\{ \matrix{ a_{\,n} \in Q = \left( { - 1} \right)^{\,n + 1} f(n) \hfill \cr \left\{ {a_{\,n} } \right\} = \left\{ {a_{\,2n} } \right\} \cup \left\{ {a_{\,2n + 1} } \right\}\quad \left| {\;\left\{ {a_{\,2n} } \right\} \cap \left\{ {a_{\,2n + 1} } \right\} = \emptyset } \right. \hfill \cr} \right. $$ and that $$ \eqalign{ & \mathop {\lim }\limits_{n\; \to \;\infty } a_{\,2n} = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - 1} \right)^{\,2n + 1} f(2n) = - \mathop {\lim }\limits_{x\; \to \;\infty } f(x) = - 2 \cr & \mathop {\lim }\limits_{n\; \to \;\infty } a_{\,2n + 1} = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - 1} \right)^{\,2n + 2} f(2n + 1) = \mathop {\lim }\limits_{x\; \to \;\infty } f(x) = 2 \cr} $$ and having demonstrated that $f(x)$ is increasing $$ f(n - 1) < f(n) < f(n + 1) $$ that means $$ \eqalign{ & - 2 < a_{\,2n} \, \le a_{\,0} = 0 \cr & 1/2 = a_{\,1} \le a_{\,2n + 1} \, < 2 \cr} $$ and therefore $$ \eqalign{ & \inf \left\{ {a_{\,n} } \right\} = \min \left( {\inf \left\{ {a_{\,2n} } \right\},\inf \left\{ {a_{\,2n + 1} } \right\}\;} \right) = - 2 \cr & \sup \left\{ {a_{\,n} } \right\} = \max \left( {\sup \left\{ {a_{\,2n} } \right\},\sup \left\{ {a_{\,2n + 1} } \right\}\;} \right) = 2 \cr} $$