Unbounded sequences diverge – Serlo

Aus Wikibooks

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[Bearbeiten]

In the chapter „Konvergenz und Divergenz beweisen“ we have already seen that the sequence diverges. We used the property that the sequence grows beyond any boundary. That is to say if we have fixed, then there exists with . Also for all with we have and thus

Infinitely many members lie outside of the -neighbourhood . Therefore cannot converge towards . If that were the case almost all members would have to be contained inside , which is not the case. Because was chosen arbitrary, the sequence 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 becomes arbitrary large. Remember the definition of an unbounded sequence:

A sequence is unbounded, if for all there are infinitely many members with .

We can use this property to prove the following theorem:

Theorem (Unbounded sequences diverge)

Let be an unbounded sequence. Then the sequence 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 arbitrary and show that the unbounded sequence cannot converge towards . For this we show that is bigger than a certain for infinitely many . We must find a lower bound of . Here we can use the inverted triangle inequality:

We know that infinitely many are larger than any fixed bound , since is unbounded. We set . Now we can proceed as we did in the example above with :

From it follows that is not contained in the -neighbourhood of . But this means that cannot be the limit of , which is precisely what we wanted to prove.

Proof (Unbounded sequences diverge)

Let be an unbounded sequence. Let be arbitrary. Since is unbounded, there are infinitely many with . For this members it follows:

These infinitely many members are outside of the interval . Hence cannot converge to . Because the choice of was arbitrary, must diverge.

Example (Geometric progression)

A generalisation of the introductory example: For the geometric progression diverges. From the Folgerungen aus der Bernoulli-Ungleichung for every there are infinitely many with . So is unbounded and consequently divergent.

Convergent sequences are bounded[Bearbeiten]

Proof by contraposition[Bearbeiten]

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:

The theorem is the following implication

Using contraposition we obtain the equivalent statement:

But this means that

If you are not sure that contraposition works just make the truth table of and compare it to that of . As an example: "If it rains (), the ground is wet ()." The contraposition is: "If the gound is not wet (), it doesn't rain ()." 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 is convergent, then there exists with for all .


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 . This sequences is bounded, but not convergent.

Alternative direct proof[Bearbeiten]

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 -definition of the limit in a proof.

Proof (Convergent sequences are bounded)

Let be a convergent sequence. Then there must be an , so that for all there exists an index with for all (this is the -definition of the limit).

We fixate (we could have chosen any other ). Now there exists an index , so that for all . All the members following are contained inside the -neighbourhood . Thus all members starting from must be smaller than :

Before there are only finitely many members, so there is a maximum . So now we have:

Altogether the elements of the sequences are bounded above by the maximum of and , so the sequences is bounded. Analogous we can show that the sequences is bounded below. Therefore the sequence is bounded.