Math for Non-Geeks/Divergence to infinity/Rules

oben In the last article, we mentioned that some rules for limit calculation carry through to improperly convergent sequences, and some don't. For instance, if a sequence (improperly) converges to and a second sequence (properly) converges to , one cannot make any statement about the convergence of their product! Find examples for sequences and as above, such that where . is bounded and diverges. is unbounded, but does not converge improperly Part 1: For instance, with and , there is , and . Part 2: For instance, with and , there is , and . Part 3: For instance, with and , there is , and . Part 4: For instance, with and , there is , and . Part 5: For instance, with and , there is , and is bounded and diverges. Part 6: For instance, with and , there is , and is unbounded, but does not converge improperly. Rules for computing limits of improperly converging sequences Which calculation rules for limits of convergent sequences can be carried over to improper convergence? The answer is: almost all of them, but only if certain conditions hold! Product rule Suppose that is a sequence with . What will happen to the product ? The case definitely causes trouble, meaning that we cannot make any statement about convergence or divergence of the product. Case 1: . Intuitively, so we expect . This assertion only needs to be mathematically proven: Let be given. Since we can find an with for all . Analogously, since there is an with for all . Whenever we therefore have So, indeed . Case 2: . Intuitively, so we expect . What we need to show for a mathematical proof is: So let again be given. Since there is an with for all . Analogously, since there is an with for all . Now, for all we have And indeed there is . Case 3: . Intuitively, for so we again make a guess . The proof could be done as the two examples above. However, this time we will vary it a bit, to make it not too boring: Let be given. Since , for each there is an with for all . We set . Then there is : , which especially includes . Since there is some with for all . Now for we have And hence . Case 4: . Here, . Prove this. We need to show: So let be given. Since , we have that for any some exists with for all . Now, set . Then : , which especially includes . Since there is also an with for all . Now, for there is And we obtain the desired result . Those four cases can also be combined into one statement. We introduce a practical extension of the real numbers: To the set , we add the elements which leads to the bigger set . Product rule for improperly convergent sequences Let be a real sequence with and let be a real sequence with . Then Sum rule Let again be a sequence with . What can we say about the limit of a sum ? If converges to a finite limit , the limit of the sum will be , as intuitively . Similarly . The critical case is , as is not well-defined. In fact, this case does not allow for any statement about convergence or divergence of the sum . As an example, For and there is . For and there is . We therefore exclude the case and consider all other cases: Case 1: . We expect . Mathematically, we need to prove: Let be given. Since there is an with for all . Analogously, since there is an with for all . Hence, for all we have And indeed . Case 2: . We also expect . Mathematically, we need to prove: Let be given. Since for each we can find an with for all . This includes the case . Hence, for all . Since there is also an with for all . Hence, for any we have And we get the desired result . Both cases can be concluded in a theorem: Sum rule for improperly convergent sequences Let be a real sequence with and be a real sequence with . Then Inversion This rule is also quite intuitive: Let be a sequence with for all and or , then should converge to and hence be a null sequence. Is this really mathematically true? Case 1: . We need to show Let be given. Since for any there is an , such that there is . Hence, there is So . Case 2: . Prove that in this case, again converges to 0. Here, we also need to show: Let be given. Since , for any there is an , such that there is . Hence, . So there is . We conclude these findings in a theorem: Inversion rule 1 for improperly convergent sequences Let be a real sequence with for all and . Then Question Does the converse of the inversion rule hold? I.e. if , may we conclude that or ? Nope! A counterexample is with . Then, , but does not go to or . However, it is true that the absolute values diverge to . The question is now: can we define a "converse of the inversion rule" which holds under more special assumptions? The sequence is not diverging to or because it keeps changing sign, so there is a subsequence of it converging to and a subsequence converging to . We can avoid this by forbidding a change of sign in . It should also not be too bad if the change of sign is allowed again on finitely many elements, since a manipulation on finitely many elements never changes convergence properties. Case 1: Let be a sequence with , all sequence elements being and all but finitely many sequence elements being positive. Then, intuitively . For a mathematical proof, we need to show that Let be given. Since is a null sequence, for we can find an with for all . Since almost all elements of are positive, there is an with for all . Therefore for all . So we get . Case 2: Let now be a sequence with , all sequence elements being but this time, almost all of them are negative. Prove that in this case . This is a kind of "intelligent copycat exercise": We take the proof above and perform some minor changes in the signs. Our aim is to show: Let be given. Since is a null sequence, for we can find an with for all . Since almost all elements of are negative, there is an with for all . Therefore , and hence for all . So we get . The converse of the inversion rule is also concluded into a theorem. Inversion rule 2 for improperly convergent sequences Let be a real sequence with for all and . Then , if for almost all , if for almost all Inversion rule In the article examples for limits, we have proven that "exponential sequences win against polynomial ones", i.e. is a null sequence for and . If , all sequence elements are positive. So the inversion rule implies Analogously, for : Quotient rule The inversion rule is an example of a quotient of sequences and with constant . Now, we generalize to quotients of any sequences and with for all . First, we consider . At this point, we exclude the cases , since is ill-defined. Let . Then, formally . To verify this mathematically, we need to show Let be given. Since there is convergence , the sequence must be bounded, i.e. there is a with for all . Now since , the inversion rule implies . So there is an with for all . Hence, there is: And we have convergence . The case and also leads to by the same argument. We conclude Quotient rule 1 for improperly convergent sequences Let and be real sequences with for all , , and . Then . Next, we let the numerator diverge as . The case again leads to the ill-defined expression and will not be considered at this point. Case 1: . Here, we assert : Let be given. Since converges to , there must be an , such that for all . Since there is also an with for all . Hence, for all , there is: So we have convergence . Case 2: with almost all being positive. Here, we assert . A mathematical proof requires showing Let be given. Since converges to and almost all elements are positive, there must be an with for all . Since there is also an with for all . So for all , there is: And again, we have convergence . Prove that for and for , with almost all being negative, there is: Case 1: . We need to prove: Let be given. Since converges to , there is an , such that for all . Since there is an with for all . Hence, for all there is: So we have convergence . Case 2: and almost all are negative. Again, we need to show: Let be given. Since converges to and almost all elements are negative, there is an with for all . Since there is an with for all . Hence, for all there is: And again, we have convergence . All 4 cases are concluded in a theorem: Quotient rule 2 for improperly convergent sequences Let and be real sequences with for all and , and suppose that . If or with almost all , then . If or with almost all , then . Direct comparison Intuitively, if is given and some "smaller" sequence diverges to , then also the "bigger" must tend to . This should still hold true if " is bigger than " almost everywhere. Mathematically, we need to show So let be given. Since there is an with for all . Since for all but finitely many there is an with for all . So indeed, . We conclude this in a theorem: Direct comparison for sequences Let be a real sequence and be another sequence with for almost all and let . Then, . Direct comparison for sequences Take the sequence . For there is In addition, . So by direct comparison, Of course, a similar statement holds true for for almost all and . Then also . This can easily be seen by applying the comparison rule to , since . unten