The minimum and maximum of a sequence

[Vali]

I have the following sequence:
$x_{n}=(-1)^{n-1}\cdot \left(2+\frac{3}{n}\right),\forall n\in \mathbb{N^{*}}$
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?

 

[joypav]

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?

Hopefully I am understanding what you are asking...
The problem is asking for the maximum and minimum of the given sequence.
You are correct that the minimum is $-\frac{7}{2}$. The max should be $5$.

You don't need to find the maximum of the subsequence $x_{2k}$ because any $x_{2k}$ will be negative, so it will surely not be the maximum of a sequence which includes positive values!

 

[Vali]

Yes, from responses, the maximum should be 5 but I want to know how to find it(let's say that I don't have the answers)
I don't want to find the maximum of $x_{2k}$.I need to find the maximum of $x_{k}$.
I mean, the $x_{2}<x_{4}<x_{6}<...<x_{2n}<...$ part is just the beginning of the $x_{k}$ terms.I want to co put together, to sort the terms of $x_{2k}$ and $x_{2k+1}$ to too see the order of $x_{k}$This way, I can see the minimum and the maximum.Those terms are the sorted terms from $x_{2k}$ and I need to continue with $x_{2k+1}$ terms so I'll get something like this:
$$x_{2}<x_{4}<...<x_{2k}<unknown \ term<another\ unknown \ term < ...<x_{2k+1}<...<x_{3}<x_{1}$$
So the max term is $x_{1}=5$ , which terms are between these subsequences?

 

[joypav]

Again.. maybe I don't quite understand what you are asking. But someone else may understand and respond! Anyways, I will try again to answer.

We have a sequence

$x_1, x_2, x_3, x_4, x_5, x_6, x_7, ...$

equal to

$5, -\frac{7}{2}, 3, -\frac{11}{4}, \frac{13}{5}, -\frac{15}{6}, ...$

We have two subsequences that are easy to spot, $x_{2k}$ which is increasing as well as always being negative, and $x_{2k+1}$ which is decreasing as well as always positive.

Now let's take a look at how we can determine the max and min of our original sequence, $x_n$.
Look at each subsequence, (as you are already doing).

Subsequence $x_{2k}$:

$x_2, x_4, x_6, ...$

or

$-\frac{7}{2}, -\frac{11}{4}, -\frac{15}{6}, ...$

This subsequence is increasing. So the first term of the subsequence must be its smallest term, $-\frac{7}{2} = -3.5$.
But how big can it increase to? To determine this value, take the limit as our n gets bigger and bigger.

$\lim_{{n}\to{\infty}} -2 - \frac{3}{n} = -2 + 0 = -2$

So this subsequence will never reach a value larger than -2.

Subsequence $x_{2k+1}$:

$x_1, x_3, x_5, ...$

or

$5, 3, \frac{13}{5}, ...$

This subsequence is decreasing. So the first term of the subsequence must be its largest term, 5.
But how small can it decrease to? To determine this value, take the limit as our n gets bigger and bigger.

$\lim_{{n}\to{\infty}} 2 + \frac{3}{n} = 2 + 0 = 2$

So this subsequence will never reach a value smaller than 2.

Conclusion:

Okay, all that work is out of the way! So let's look at the values we've determined.

One subsequence has smallest value $-3.5$, and values no bigger than $-2$.
The other has largest value $5$, and values no smaller than $2$.

Looking at these four values we've determined, we can now decide what our max and min is. When we put those two subsequences back together we get our originally given sequence.

Our smallest value for $x_{2k}$ is -3.5 and our smallest value for $x_{2k+1}$ is 2. Which is the smaller? -3.5
Then -3.5 is the minimum of the whole sequence.

Our largest value for $x_{2k}$ is -2 and our smallest value for $x_{2k+1}$ is 5. Which is the larger? 5
Then 5 is the maximum of the whole sequence.

- - - Updated - - -

$$x_{2}<x_{4}<...<x_{2k}<unknown \ term<another\ unknown \ term < ...<x_{2k+1}<...<x_{3}<x_{1}$$
So the max term is $x_{1}=5$ , which terms are between these subsequences?

Additionally... if you put them in order like this, the values in between the two sequences are infinite.
Your even indices get closer and closer to -2 and you odd indices get closer and closer to 2.

There are infinitely many values in the sequence. However, this does not affect the max and min, which your string of inequalities nicely shows. $x_2$ is the smallest and $x_1$ is the biggest. No need to calculate every single thing between them!

 

[Vali]

Great answer!I understood perfectly :)
Thanks a lot for your help!

 

[HOI]

Note that "A" is about monotonicity of the sequence. It say nothing about subsequences. Since the sign alternates, this sequence is clearly NOT monotonic.