Math for Non-Geeks/Unbounded sequences diverge

oben In this chapter, we will see that unbounded sequences must diverge. From this, we can deduce 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 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 of lie outside the -neighbourhood . Therefore cannot converge towards . If that were the case, almost all members of would have to be contained inside , which is not the case. Because was chosen arbitrarily, 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 arbitrarily 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: 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 sequence is unbounded, this immediately implies that it is divergent. ja As in the argument from above, we choose arbitrarily 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 with . But this means that cannot be the limit of , which is precisely what we wanted to prove. Let be an unbounded sequence. Let be arbitrary. Since is unbounded, there are infinitely many with . For these members, it follows: These infinitely many members are outside the interval . Hence cannot converge to . Because the choice of was arbitrary, must diverge. Geometric progression A generalisation of the introductory example: For the geometric progression diverges. From one of the consequences of Bernoulli's inequality, for every there are infinitely many with . So 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 deduce 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 ground is not wet (), it doesn't rain ()." These two statements are logically equivalent. So by contraposition the following theorem is true, which we will need to prove further results later on: 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 need not be convergent. And a divergent sequence need not be unbounded. As a counterexample, consider . This sequence is bounded, but not convergent. Alternative direct proof We also want to show an alternative, direct way of proving that a convergent sequence is bounded. This proof is often given in other textbooks. It shows how one can use the -definition of the limit in a 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 fix (we could have chosen any other ). Now there exists an index , which we may choose to satisfy , 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 sequence are bounded above by the maximum of and , so the sequence is bounded above. Analogously, we can show that the sequence is bounded below. Therefore, the sequence is bounded. unten