# Bounded sequences

Let $(a_n)_{n\in\mathbb{N}}$ be a sequence and $M$ a real number. We say that the sequence is bounded from above by $M$ if all the terms are smaller than $M$, that is to say, if $$a_n \leq M$$ for all $n$.

Similarly, we say that the sequence is bounded from below by $M$ if all the terms are greater than $M$, that is to say, if $$a_n \geq M$$ for all $n$. We call $M$ the upper, and lower bound of the sequence respectively.

Let's consider the sequence $a_n=\dfrac{1}{n}$. We have $0 < \dfrac{1}{n} \leq 1$ for all $n$. Therefore, the sequence is bounded from above by $1$ and bounded from below by $0$. We observe how the upper bound is part of the sequence, $a_1=1$ and, therefore, it is the best possible bound. But the lower bound does not belong to the set. This might make us think that it is not the best possible bound. Let's verify that it is not the case:

Let's suppose that $M$ is the best non zero lower bound. Since $M > 0$ we can consider a point $n$ greater than $\dfrac{1}{M}$. Then $\dfrac{1}{n} < M$ which contradicts our hypothesis on $M$.

Anyway, in many cases it is enough to find a bound even though it is not the best possible one. In fact, if a sequence is bounded from above by $M$ then it is also bounded from above by all numbers greater than $M$.

As with the classification of monotonous sequences, this one cannot either be taken as a general classification since a sequence does not necessarily admit any bound.

Let's consider the sequence $a_n=n$, which is bounded from below, for example, by $0$ but does not admit any upper bound since for all $M$ we can always find a greater $n$.

To know whether a sequence admits an upper or a lower bound is not usually a simple problem either. Therefore, it is even more difficult to find a bound, even knowing that the sequence is bounded.

In the case of monotonous sequences, the first term serves us as a bound. If we have an increasing sequence then the first term is a lower bound of the sequence. And if the sequence is decreasing then the first term is an upper bound.

Another criterion to verify whether a sequence admits a bound is to verify if all the terms of a sequence are positive, or negative, in which case the sequence would be, respectively, bounded from below by $0$ or bounded from above by $0$.

## Calculation using functions

When a sequence is given by the general term, we can verify if a sequence is monotonous or bounded from the function that defines the general term. In this case, the properties of the function are also valid for the sequence. More specifically;

If the function is monotonous for values greater than $1$, then the sequence is monotonous too. It can even happen that the sequence is definitely monotonous even though the function is not.

For example, let's consider the sequence $\Big(\Big(n-\dfrac{3}{2}\Big)^2\Big)_{n\in\mathbb{N}}$. The function $f(x)=\Big(x-\dfrac{3}{2}\Big)^2$ is not increasing for values greater than $1$, for example $$f(1)=\dfrac{1}{4} > 0 =f\Big(\dfrac{3}{2}\Big)$$ But we can verify that the sequence is increasing since $$\Big(n-\dfrac{3}{2}\Big)^2 \leq \Big(n-\dfrac{1}{2}\Big)^2$$

If the function is bounded from above, or below, for values greater than 1 then the sequence is also bounded from above, or below, respectively.

These results allow the use of the differential calculus methods for our calculations in sequences. Essentially, the calculation of the monotony is interesting from the derivative of the general term.

As a last comment, we can think about whether the reciprocal of the previous results is true or not.In fact, we can ask ourselves if the results obtained for the sequence are also valid for the function. The answer to this question is negative in general. For example,

For example, let's consider the function $f(x)=x\cdot\sin(2\pi x)$. We calculate the corresponding sequence. We evaluate the function in an integer $n$; $$f(n)=n\cdot\sin(2\pi n)=n\cdot\sin(2\pi)=0$$ Namely, it defines the constant function. And this is increasing, decreasing and upper and bounded from below.

Now we are going to see that the function is not any of those. We evaluate the function in the points $n+\dfrac{1}{4}$ for $n$ integer $$f\Big(n+\dfrac{1}{4}\Big)=\Big(n+\dfrac{1}{4}\Big)\cdot\sin(2\pi n+\dfrac{2\pi}{4})=\Big(n+\dfrac{1}{4}\Big)\cdot\sin\Big(\dfrac{\pi}{2}\Big)=\Big(n+\dfrac{1}{4}\Big)$$

We see therefore that the function is not already bounded from above since $f\Big(n+\dfrac{1}{4}\Big)=n+\dfrac{1}{4}$ cannot be bounded from above for any $n$.

We evaluate now the function at the points $n+\dfrac{3}{4}$ for $n$ integer; $$f\Big(n+\dfrac{3}{4}\Big)=\Big(n+\dfrac{3}{4}\Big)\cdot\sin(2\pi n+2\pi\dfrac{3}{4})=\Big(n+\dfrac{3}{4}\Big)\cdot\sin\Big(\dfrac{3\pi}{2}\Big)=-\Big(n+\dfrac{3}{4}\Big)$$

In the same way we establish that the function is not bounded from below since $$f\Big(n+\dfrac{3}{4}\Big)=-\Big(n+\dfrac{3}{4}\Big)$$ cannot be bounded from below for any $n\in\mathbb{N}.$

Also, since the values $n+\dfrac{1}{4}$ and $n+\dfrac{3}{4}$ are inserted into one another, it turns out that the function cannot be monotonous since in the first ones the function is positive and in the second ones it is negative.