Divergence to infinity: rules – Serlo

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

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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

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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

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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

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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

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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

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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  .