Divergence to infinity: rules – Serlo

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In the last article, we mentioned that some rules for limit calculation carry through to improper 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!

Exercise

Find examples for sequences and as above, such that

  1. where
  2. is bounded and diverges
  3. is unbounded, but does not converge improperly

Solution

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[Bearbeiten]

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[Bearbeiten]

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, 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, .

Exercise

Prove this.

Solution

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 concluded 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 .

Theorem (Product rule for improperly convergent sequences)

Let be a real sequence with and a real sequence with . Then

Sum rule[Bearbeiten]

Let again be a sequence with . What can we say about the limit of a sum  ? For finite , the limit will stay unchanged, 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:

Theorem (Sum rule for improperly convergent sequences)

Let be a real sequence with and be a real sequence with . Then

Inversion[Bearbeiten]

This rule is also quite intuitive: Let be a sequence with for all and or , then formally converges to and should 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: .

Exercise

Prove that in this case, again converges to 0.

Solution

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:

Theorem (Inversion rule 1 for improperly convergent series)

Let be a real sequence with for all and . Then

Question: Does the converse of the inversion rule, i.e. if we may imply 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 presign, so there is a subsequence of it converging to and a subsequence converging to . We can avoid this by forbidding a change of presign in . It should also not be too bad if the change of presign 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 v , all sequence elements being but this time, almost all of them are negative.

Exercise

Prove, that in this case .

Solution

This is a kind of "intelligent copycat exercise": We take the proof above and perform some minor changes in the presigns. 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.

Theorem (Inversion rule 2 for improperly convergent series)

Let be a real sequence with for all and . Then

  • , if for almost all
  • , if for almost all

Example (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 non-negative. So the inversion rule implies

Analogously, for :

Quotient rule[Bearbeiten]

The inversion rule is an example of a quotient of sequences and wit 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

Theorem (Quotient rule 1 for improperly convergent sequences)

Let and be real sequences with and . Then .

Next, we let the enumerator diverge as . The case again leads to the ill-defined expression and will not be not 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 .

Exercise

Prove that for and for , with almost all being negative, there is:

Solution

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

Theorem (Quotient rule 1 for improperly convergent sequences)

Let and be real sequences with .

  • If or with almost all , then .
  • If or with almost all , then .

Direct comparison[Bearbeiten]

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:

Theorem (Direct comparison for sequences)

Let be a real sequence and be another sequence with for almost all and let . Then, .

Example (Direct comparison for sequences)

Take the sequence . For there is

In addition . So by direct comparison,

Of course, a similar statement holds true for and . Then also . This can easily seen by considering the sequences and .