# Monotonous sequences

The terms of a sequence have, in principle, neither any relation nor any order.When looking for the general term of a sequence we have studied some relations among the terms of the sequences and now we will focus on the order relations.

Considering a sequence $(a_n)_{n\in\mathbb{N}}$ we say that it is increasing if any element of the sequence is smaller than the next ones, that is, if $a_n \leq a_{n+1}$ for all $n$.

Similarly, we say that it is decreasing if any element of the sequence is greater than the next ones, $a_n \geq a_{n+1}$ for all $n$.

For a sequence that is as much increasing as it is decreasing, we say that it is constant since then $a_n=a_{n+1}$, all the terms are equal.

Considering an increasing sequence we say that it is strictly increasing if $a_n < a_{n+1}$. And, similarly, a decreasing sequence is strictly decreasing if $a_n > a_{n+1}$.

The corresponding sequence to $a_n=n$ is clearly increasing since $n < n+1$. In fact it is strictly increasing.

As an example of a decreasing sequence we can consider the sequence with the general term $a_n=\dfrac{1}{n}$. We observe that $\dfrac{1}{n} > \dfrac{1}{n+1}$ and the sequence is strictly decreasing.

A constant sequence has the form $a_n=c$ where $c$ is any number. Any constant sequence has this form.

This classification must not be taken as generic since any given sequence is not always increasing or decreasing.

If a sequence is increasing or decreasing we will say that it is monotonous. If, furthermore, it is a strictly increasing or a strictly decreasing sequence we call it strictly monotonous.

Given a sequence, to verify whether it is increasing or decreasing is not always a simple problem. To prove these properties, the ovious way is also the most useful in most cases. We mean to outline the inequality of the property we may want to demonstrate and via all the necessary calculations to verify that it is true for all n. In this context, to verify if the sequence is strictly monotonous the already made calculations can be reused.

Let's see some more complete examples:

We consider the sequence $a=(n^2-3n)_{n\in\mathbb{N}}$. We are going to see that the sequence is increasing. Namely, we want to verify that $$n^2-3n \leq (n+1)^2-3(n+1)$$ Expanding the squares we have $$n^2-3n \leq n^2-n-2$$ and therefore $2\leq2n$, which is true for all $n\geq1$ and the sequence is increasing.

We can now verify if the sequence is strictly increasing. Repeating the same calculations from the inequality $n^2-3n < (n+1)^2-3(n+1)$ we would obtain $2 < 2n$, which is only true for $n > 1$. Namely, for $n=1$ the inequality is not satisfied. Taking a look at the first values of the sequence we see which phenomenon takes place: $$a=(-2,-2,0,4,10,18,\ldots)$$

Let's now consider the sequence $b=\Big(\dfrac{n+1}{n^2}\Big)_{n\in\mathbb{N}}$. Let's see that it is decreasing. We must verify that $$\dfrac{n+1}{n^2} \geq \dfrac{(n+1)+1}{(n+1)^2}$$ Likewise, $$(n+1)^2\cdot(n+1)\geq (n+2)\cdot n^2$$ By expanding, we have $$n^3+3n^2+3n+1 \geq n^3+2n^2$$ And it is necessary to be satisfied that $$n^2+3n+1 \geq 0$$ To verify this type of inequality it is sufficient to verify that there is no an integer between the solutions of the polynomial equal to zero. In our case the solutions of $n^2+3n+1 =0$ are approximately $-2.62$ and $-0.38$. And, according to the previous comment, the inequality is satisifed and the sequence is decreasing.

To verify if it is strictly decreasing, we can repeat the calculations until obtaining $n^2+3n+1 > 0$ from the condition $$\dfrac{n+1}{n^2} > \dfrac{(n+1)+1}{(n+1)^2}$$ To verify that the sequence is strictly decreasing it is enpough to see that the roots obtained previously are not natural numbers.