As with functions, we want to study how our sequence evolves: does value increase, decrease, neither or both? Let's start with the definition of these variations:

A sequence $$u_ {n}$$ is increasing if for any natural number n:
$$U_{n+1} \geqslant U_{n}$$
A sequence $$u_ {n}$$ is strictly increasing if for any natural number n:
$$U_{n+1} > U_{n}$$
A sequence $$u_ {n}$$ is constant if for any natural number n:
$$U_{n+1} = U_{n}$$
Note: for the decrease, just reverse the signs.

When we are asked "to study the monotony of the sequence" , we have to find out what situation we are in.

The sequence $$u_ {n}$$ is (strictly) monotone if and only if it is (strictly) increasing or (strictly) decreasing.

So, how do we find the relationship we're in?

For this, it is usually sufficient to calculate $$U_{n+1} - U_{n}$$:

if $$U_{n+1}-U_{n} \geqslant 0$$ for all $$n \in \mathbb{N}$$, the sequence $$U_{n}$$ is increasing
if $$U_{n+1}-U_{n} \leqslant 0$$ for all $$n \in \mathbb{N}$$, the sequence $$U_{n}$$ is decreasing
if $$U_{n+1}-U_{n} = 0$$ for all $$n \in \mathbb{N}$$, the sequence $$U_{n}$$ is constant

A sequence can be neither increasing nor decreasing (therefore neither monotonic), nor constant. This is the case, for example, of the sequence defined by $$U_{n} = (-1)^{n}$$ whose terms are successively : 1; -1; 1; -1; 1; ...

Note: the study of variations for arithmetic sequences and geometric sequences are discussed at greater length on their respective pages.

## $$U_{n+1} = U_n^2+U_n$$

Resolution

Consider the sequence $$un$$ defined by recurrence by:

$$U_0 = 1$$
$$U_{n+1} = U_n^2+U_n$$

We calculate the difference: $$U_{n+1} - U_n = U_n^2 + U_n - U_n = U_n^2$$ We know that the function x² is always positive or zero, so: $$U_n^2\geq0 \quad et \quad U_{n+1}-U_n\geq0$$
We can therefore conclude that the sequence $$U_n$$ is increasing.

## $$U_n=\dfrac1n$$

Resolution

Consider the sequence $$un$$ explicitly defined by:

$$U_n=\dfrac1n$$ for all n > 0

We calculate the difference: $$U_{n+1} - U_n = \dfrac{1}{n+1} - \dfrac1n = \dfrac{n - (n+1)}{n(n+1)} = \dfrac{-1}{n(n+1)}$$ However, for any non-zero natural number n, we can deduce that: $$\dfrac{-1}{n(n+1)}\leq0 \quad et \quad U_{n+1}-U_n\leq0$$
We can therefore conclude that the sequence $$U_n$$ is decreasing.

## $$U_{n+1} = U_n - 1$$

Resolution

Consider the sequence $$un$$ defined by recurrence by:

$$U_0 = 13$$
$$U_{n+1} = U_n - 1$$

We calculate the difference: $$U_{n+1} - U_n = U_n - 1 - U_n = -1$$ Then, for all n : $$-1 \lt 0 \quad et \quad U_{n+1}-U_n \lt 0$$
We can therefore conclude that the sequence $$U_n$$ is strictly decreasing.

Tada! We have seen that to study the direction of variation of a sequence, we simply need to calculate the difference between the term "n + 1" and its predecessor "n". We are now able to approach the notions of limit and convergence, or else put into practice what we have just seen with arithmetic sequences and geometric sequences.