- #1

Vali

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$x_{n}=(-1)^{n-1}\cdot \left(2+\frac{3}{n}\right),\forall n\in \mathbb{N^{*}}$

I need to choose from:

A. $x_{n}$ is a monotonic sequence

B. $x_{n}$ limit is $2$

C. $x_{n}$ minimum is $-\frac{7}{2}$ and the maximum is $5$

D. $x_{n}$ minimum is $-2$ and the maximum is $2$

First, I rewrote the sequence:

$$x_{n}=\begin{cases} -2-\frac{3}{n} & \text{ if } n \ is \ even \\ 2+\frac{3}{n} & \text{ if } n \ is \ odd \end{cases}$$

I checked the $x_{n}$ monotony with n even and n odd like:

$x_{2n+1}-x_{2n}=2+\frac{3}{2n+1}+2+\frac{3}{2n}>0$ so the $x_{2n}$ subsequence is increasing

$x_{2n+2}-x_{2n+1}=-2-\frac{3}{2n+2}-2-\frac{3}{2n+1}<0$ so the $x_{2n+1}$ subsequence is decreasing

Now, how should I write the terms to see the minimum and maximum?

For now, I wrote

$x_{2}<x_{4}<x_{6}<...<x_{2n}<...$ how to continue with the decreasing subsequence to see the maximum?

The minimum is $x_{2}=-\frac{7}{2}$ ,right?