Extrema of this sequence 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
$$\left|\left(2-\frac{n+2}{2^n}\right)\right|&lt;2$$

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:

$$a_{2k}&lt;2 \Rightarrow 1\left(2-\frac{2k+2}{2^{2k}}\right)&lt;2, \text{ true for } k&gt;-1$$

$$a_{2k-1}&gt;-2 \Rightarrow -1\left(2-\frac{2k+1}{2^{2k-1}}\right)&gt;-2, \text{ true for } k&gt;-\frac1{2}$$

So I know that $a_n$ is bounded.

I tried to demonstrate that $\sup=2$ and $\inf=-2$ by doing the following:

$$a_{2k}&gt;2-\varepsilon \Rightarrow 1\left(2-\frac{2k+2}{2^{2k}}\right)&gt;2-\varepsilon$$

$$a_{2k-1}&lt;-2+\varepsilon \Rightarrow -1\left(2-\frac{2k+1}{2^{2k-1}}\right)&lt;-2+\varepsilon,$$
but I can't solve those inequalities. Is this one the right approach?

EDIT 

Another attempt: (Thanks to G Cab's answer)

$$-b_n&lt;a_n&lt;b_n$$

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 

$$-2&lt;-b_n&lt;a_n&lt;b_n&lt;2$$

and $\sup(a_n)=2$ and $\inf(a_n)=-2$
Note that
$$
\eqalign{
  &amp; f(x) = 2 - {{x + 2} \over {2^{\,x} }}\quad  \Rightarrow \quad y(0) = 0  \cr 
  &amp; f'(x) = {{\ln 2\left( {x + 2} \right) - 1} \over {2^{\,x} }}  \cr 
  &amp; 0 &lt; f'(x) = \quad  \Rightarrow \quad {1 \over {\ln 2}} - 2 \approx  - 0.557 &lt; x \cr} 
$$
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
$$
\eqalign{
  &amp; \mathop {\lim }\limits_{n\; \to \;\infty } a_{\,2n}  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - 1} \right)^{\,2n + 1} f(2n) =  - \mathop {\lim }\limits_{x\; \to \;\infty } f(x) =  - 2  \cr 
  &amp; \mathop {\lim }\limits_{n\; \to \;\infty } a_{\,2n + 1}  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - 1} \right)^{\,2n + 2} f(2n + 1) = \mathop {\lim }\limits_{x\; \to \;\infty } f(x) = 2 \cr} 
$$
and having demonstrated that $f(x)$ is increasing
$$
f(n - 1) &lt; f(n) &lt; f(n + 1)
$$
that means
$$
\eqalign{
  &amp;  - 2 &lt; a_{\,2n} \, \le a_{\,0}  =  0  \cr 
  &amp; 1/2 = a_{\,1}  \le a_{\,2n + 1} \, &lt; 2 \cr} 
$$
and therefore
$$
\eqalign{
  &amp; \inf \left\{ {a_{\,n} } \right\} = \min \left( {\inf \left\{ {a_{\,2n} } \right\},\inf \left\{ {a_{\,2n + 1} } \right\}\;} \right) =  - 2  \cr 
  &amp; \sup \left\{ {a_{\,n} } \right\} = \max \left( {\sup \left\{ {a_{\,2n} } \right\},\sup \left\{ {a_{\,2n + 1} } \right\}\;} \right) = 2 \cr} 
$$
I think, you are right because $$\lim_{n\rightarrow+\infty}a_{2n}=-2$$ and

$$\lim_{n\rightarrow+\infty}a_{2n-1}=2$$ and use your work.