# Math for Non-Geeks/Unbounded sequences diverge

In this chapter we will see that unbounded sequences must diverge. From this we can follow that convergent sequences must be bounded.

## Unbounded sequences diverge

In the chapter How to prove convergence and divergence we have already seen that the sequence $\left(2^{n}\right)_{n\in \mathbb {N} }$ diverges. We used the property that the sequence grows beyond any boundary. That is to say if we have $a\in \mathbb {R}$ fixed, then there exists $N\in \mathbb {N}$ with $2^{N}\geq a+1$. Also for all $n\in \mathbb {N}$ with $n\geq N$ we have $2^{n}\geq a+1$ and thus
${\begin{aligned}|2^{n}-a|&=2^{n}-a\\&\geq (a+1)-a=1\end{aligned}}$
Infinitely many members $\left(2^{n}\right)_{n\in \mathbb {N} }$ lie outside of the $\epsilon$-neighbourhood $(a-1;a+1)$. Therefore $\left(2^{n}\right)_{n\in \mathbb {N} }$ cannot converge towards $a$. If that were the case almost all members $\left(2^{n}\right)_{n\in \mathbb {N} }$ would have to be contained inside $(a-1;a+1)$, which is not the case. Because $a$ was chosen arbitrary, the sequence $\left(2^{n}\right)_{n\in \mathbb {N} }$ cannot have a limit and is therefore divergent.
We can extend this argument to any sequence that is unbounded, since we only used the property that $\left(2^{n}\right)_{n\in \mathbb {N} }$ becomes arbitrary large. Remember the definition of an unbounded sequence:
We can use this property to prove the following theorem:
Theorem (Unbounded sequences diverge)
Let $(a_{n})_{n\in \mathbb {N} }$ be an unbounded sequence. Then the sequence $(a_{n})_{n\in \mathbb {N} }$ is divergent.
With this theorem we can prove that a sequence is divergent. If we can show that a sequences is unbounded, this immediately implies that it is divergent.
How to get to the proof? (Unbounded sequences diverge)
As in the argument from above we choose $a$ arbitrary and show that the unbounded sequence $(a_{n})_{n\in \mathbb {N} }$ cannot converge towards $a$. For this we show that $|a_{n}-a|$ is bigger than a certain $\epsilon >0$ for infinitely many $n\in \mathbb {N}$. We must find a lower bound of $|a_{n}-a|$. Here we can use the inverted triangle inequality:
$|a_{n}-a|\geq ||a_{n}|-|a||$
We know that infinitely many $|a_{n}|$ are larger than any fixed bound $S\geq 0$, since $(a_{n})_{n\in \mathbb {N} }$ is unbounded. We set $S=|a|+1$. Now we can proceed as we did in the example above with $\left(2^{n}\right)_{n\in \mathbb {N} }$:
${\begin{aligned}|a_{n}-a|&\geq ||a_{n}|-|a||\\[0.3em]&{\color {OliveGreen}\left\downarrow \ |a_{n}|\geq |a|+1\geq |a|\right.}\\[0.1em]&\geq |a_{n}|-|a|\\[0.3em]&{\color {OliveGreen}\left\downarrow \ |a_{n}|\geq |a|+1\right.}\\[0.1em]&\geq (|a|+1)-|a|=1\end{aligned}}$
From $|a_{n}-a|\geq 1$ it follows that $a_{n}$ is not contained in the $\epsilon$-neighbourhood of $(a-1;a+1)$. But this means that $a$ cannot be the limit of $(a_{n})_{n\in \mathbb {N} }$, which is precisely what we wanted to prove.
Proof (Unbounded sequences diverge)
Let $(a_{n})_{n\in \mathbb {N} }$ be an unbounded sequence. Let $a\in \mathbb {R}$ be arbitrary. Since $(a_{n})_{n\in \mathbb {N} }$ is unbounded, there are infinitely many $a_{n}$ with $|a_{n}|\geq |a|+1$. For this members it follows:
${\begin{aligned}&|a_{n}-a|\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{inverted triangle inequality}}\right.}\\[0.3em]&\geq ||a_{n}|-|a||\\[0.3em]&{\color {OliveGreen}\left\downarrow \ |a_{n}|\geq |a|+1\geq |a|\right.}\\[0.3em]&\geq |a_{n}|-|a|\\[0.3em]&{\color {OliveGreen}\left\downarrow \ |a_{n}|\geq |a|+1\right.}\\[0.3em]&\geq (|a|+1)-|a|=1\end{aligned}}$
These infinitely many members $a_{n}$ are outside of the interval $(a-1;a+1)$. Hence $(a_{n})_{n\in \mathbb {N} }$ cannot converge to $a$. Because the choice of $a$ was arbitrary, $(a_{n})_{n\in \mathbb {N} }$ must diverge.
Example (Geometric progression)
A generalisation of the introductory example: For $|q|>1$ the geometric progression $\left(q^{n}\right)_{n\in \mathbb {N} }$ diverges. From one of the consequences of Bernoulli's inequality, for every $S>0$ there are infinitely many $n\in \mathbb {N}$ with $|q^{n}|=|q|^{n}>S$. So $\left(q^{n}\right)_{n\in \mathbb {N} }$ is unbounded and consequently divergent.

## Convergent sequences are bounded
### Proof by contraposition

The above theorem tells us that unbounded sequences are divergent. With the aid of logical contraposition, we can follow that convergent sequences must be bounded. The principle of contraposition is:
$(A\implies B)\iff (\neg B\implies \neg A)$
The theorem is the following implication
$(a_{n})_{n\in \mathbb {N} }{\text{ is unbounded}}\implies (a_{n})_{n\in \mathbb {N} }{\text{ diverges}}$
Using contraposition we obtain the equivalent statement:
$\neg {\big(a_{n})_{n\in \mathbb {N} }{\text{ diverges}}{\big )}\implies \neg {\big(a_{n})_{n\in \mathbb {N} }{\text{ is unbounded}}{\big )}$
But this means that
$(a_{n})_{n\in \mathbb {N} }{\text{ converges}}\implies (a_{n})_{n\in \mathbb {N} }{\text{ is bounded}}$
If you are not sure that contraposition works just make the truth table of $(A\implies B)$ and compare it to that of $(\neg B\implies \neg A)$. As an example: "If it rains ($A$), the ground is wet ($B$)." The contraposition is: "If the gound is not wet ($\neg B$), it doesn't rain ($\neg A$)." This two statements are logically equivalent.
So by contraposition the following theorem is true, which we will need to prove further results later on:
Theorem (Convergent sequences are bounded)
Every convergent sequence is bounded. Thus if a sequence $(a_{n})_{n\in \mathbb {N} }$ is convergent, then there exists $S\geq 0$ with $|a_{n}|\leq S$ for all $n\in \mathbb {N}$.
Hint
The converse of the above statement is not true. This means: A bounded sequence must not be convergent. And a divergent sequence must not be unbounded.
As a counterexample consider $((-1)^{n})_{n\in \mathbb {N} }$. This sequences is bounded, but not convergent.

### Alternative direct proof

We also want to show an alternative, direct way of proving that a convergent sequences is bounded. This proof is often given in other textbooks. It shows how one can use the $\epsilon$-definition of the limit in a proof.
Proof (Convergent sequences are bounded)
Let $(a_{n})_{n\in \mathbb {N} }$ be a convergent sequence. Then there must be an $a\in \mathbb {R}$, so that for all $\epsilon >0$ there exists an index $N\in \mathbb {N}$ with $|a_{n}-a|<\epsilon$ for all $n\geq N$ (this is the $\epsilon$-definition of the limit).
We fixate $\epsilon =1$ (we could have chosen any other $\epsilon >0$). Now there exists an index $N\in \mathbb {N}$, so that $|a_{n}-a|<\epsilon =1$ for all $n\geq N$. All the members following $a_{N}$ are contained inside the $\epsilon$-neighbourhood $(a-1;a+1)$. Thus all members starting from $a_{N}$ must be smaller than $a+1$:
$(a_{n})_{n\in \mathbb {N} }=(a_{1},\,a_{2},\,a_{3},\ldots ,\,{\color {OliveGreen}\underbrace {a_{N},\,a_{N+1},\,a_{N+2},\,a_{N+3},\ldots } _{{\text{These elements are smaller than }}a+1}})$
Before $a_{N}$ there are only finitely many members, so there is a maximum $M=\max\{a_{1},\,a_{2},\,a_{3},\ldots ,\,a_{N-1}\}$. So now we have:
$(a_{n})_{n\in \mathbb {N} }=({\color {Blue}\underbrace {a_{1},\,a_{2},\,a_{3},\ldots ,\,} _{{\text{These elements are less or equal to }}M\quad }}{\color {OliveGreen}\underbrace {a_{N},\,a_{N+1},\,a_{N+2},\,a_{N+3},\ldots } _{\quad {\text{These elements are less than }}a+1}})$
Altogether the elements of the sequences are bounded above by the maximum of $M$ and $a+1$, so the sequences is bounded. Analogous we can show that the sequences is bounded below. Therefore the sequence $(a_{n})_{n\in \mathbb {N} }$ is bounded.