Limit theorems – Serlo

Limit proofs using the -definition are quite laborious. In this chapter we will study some limit theorems that simplifies matters.

Limit theorems Bearbeiten

The limit theorems are the following:

Theorem (Limit theorems)

Let   and   be two convergent sequences with   and  . Let also   be arbitrary. Then we have

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  •  
  •   for all  
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If furthermore   and   for all  , then we also have

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For   and   for all  :

 

Warning

These rules can only be applied if the respective sequences converge. As soon as one of the involved sequences diverge, we cannot use the rules.

Also note that   and   are not real numbers and therefore no valid limits. If, for instance,  , then   is divergent and we cannot use the limit theorems.

Monotonicity rule: limit estimates Bearbeiten

We also have the following monotonicity rule, which can be used to estimate limits:

Theorem (Monotonicity rule)

Le   and   be two convergent sequences. If   for almost all  , then the limits satisfy the inequality

 

Example: Computing the limit of a sequence Bearbeiten

Consider the following sequence

 

This sequence is convergent. A proof using the  -definition would be rather complicated. Fortunately we can break the whole sequence apart into individual sequences where the convergence and limits are known. For example  . Using the limit theorems we can compute the limit step by step:

 

In this manner we can show that   is convergent and the limit is  . Unfortunately this derivation is flawed: We are applying the limit theorems before we showed that the individual sequences are convergent. That those sequences do indeed converges becomes clear only after we have already employed the limit theorems. Therefore the above is not a valid proof. Instead we could proceed as follows:

 

We start with the sequences of which we know that they converge. By applying the limit theorems in reverse order we can derive the convergence and limit of the original sequence  . The symbol   is the logical conjunction, which should be read as "and".

Writing the proof in the above way is time-consuming and no fun. Most of the time we prefer writing down the first version. We use the limit theorems even though we don't know if the sequences converge. We must argue afterwards, that it was okay to use the limit theorems in the first place. But this is the case, because at the end, everything converges. Since the last steps worked, we were allowed to do the steps before. So if we want to write the proof like in the first version, we need to make sure at the end to give a justification why applying the limit theorems was a valid thing to do.

Problems with divergent sequences Bearbeiten

As stated many times, we cannot use the limit theorems if one of the partial sequences diverges. If we forget this, we can quickly get nonsensical results:

 

Question: What is wrong in the above argumentation?

  is not a real number and thus   is a divergent sequence with  . A sequence is convergent only if the limit is a real number. The product rule   cannot be applied.

Proof of limit theorems Bearbeiten

Absolute value rule Bearbeiten

Theorem (Absolute value limit rule)

Let   be a convergent sequence with limit  . Then  .

How to get to the proof? (Absolute value limit rule)

If   then the distance   becomes arbitrary small. We need to show that   also becomes arbitrary small. In the chapter Absolute value we proved the following inequality:

 

Thus we have

 

This means that if   is smaller than  , by transitivity this is true also for  . We can use this in our convergence proof. Let  . We need to find  , so that   for all  . Because of   we know that there is  , so that   for all  .

But as we have seen, if   then also  . So we can set  . Since   is true for all  , it follows that   is true for all  .

Proof (Absolute value limit rule)

Let   be arbitrary. Because   converges to  , there is   with   for all  . From the inverted triangle inequality   it follows

 

for all  .


Inversion of the absolute value rule Bearbeiten

If   is a null sequence, then the inversion of the absolute value rule holds, as well. From   , we have  :

Theorem

Let   be a sequence. If   , then   converges to zero, i.e.  .

Proof

Since   , we have that

For every   there is an   with   for all  .

Now,  . Therefore, we also have that

For every   there is an   with   for all  .

Or in other words,  .

This inversion only holds for null sequences. For a general sequence, it may not hold. An example is the divergent sequence  . The absolute value sequence   converges. So   but  . The latter limit even does not exist.

The sum rule Bearbeiten

Theorem (Limit theorems for sums)

Let   be a sequence converging to   and   a sequence converging to  . Then, the sequence   also converges with  .

How to get to the proof? (Limit theorems for sums)

We need to prove that   gets arbitrarily small. What we can use is that   and   get arbitrarily small. That means, we need to construct an upper bound for   which uses   and/ or   . The trick is to use the triangle inequality in  :

 

Since   and   get arbitrarily small, this estimate should suffice. It remains to establish the bounds in terms of  . This is done by bounding each of the two summands against   : When we get   and   , then

 

We know that for some   there is   for all   . Analogously, there is some   with   for all  . For the proof, we need   and   simultaneously. So both   and   should hold for all   with some suitable   . The smallest suitable   is then given by   : From   , we get   and  .

Proof (Limit theorems for sums)

Let   be arbitrary. Then, there is an   with   for all  , since   . Analogously, since   there is an   with   for all  . We choose  . Let   be arbitrary. Then,

 

The factor rule Bearbeiten

Theorem (Factor rule for limits)

Let   be arbitrary and   a converging sequence with limit  . Then, the sequence   converges, as well with  .

How to get to the proof? (Factor rule for limits)

In order to prove   , we need to establish a bound   for almost all   . This "to-be-achieved"-inequality can be equivalently reformulated:

 

It is not allowed to directly divide by   , since there might be   . However, we can easily treat the case   : there, we need to show   . And since   , we trivially have  . So the assumption is established within case   . In the case   , we can divide by  :

 

Since   converges to   , there is an  , such that   for all  .

Proof (Factor rule for limits)

Let   be arbitrary. in addition, let   and   a sequence converging to  .

Fall 1:  

There is   so we have convergence

 

Fall 2:  

Choose   such that   for all   . Such an   can be found, since   converges to   . Then,

 

this shown the assumption   .

The product rule Bearbeiten

Theorem (Product rule for limits)

Let   be a sequence converging to   and   a sequence converging to  . Then, the product sequence   converges as well with  .

Proof (Product rule for limits)

Let   be arbitrary.

We need to show that   for all   , where we need to choose   depending on   . What we can use is that   and   get arbitrarily small, since   converges to   and   converges to   .This requires a deliberate reformulation of   , such that we get an upper bound including   and   . The trick is to add a "smart zero" in the form of  :

 

If we can get the two summands below   for all  , then we are done.

Bounding the second summand

The sequence   converges to   and by the factor rule with   we have  . Hence,   by the sum rule, so there is an   such that for all   we have  .

Bounding the first summand

This term is a bit more complicated, since the factor   is not constant. However, we can bound it from above by a constant:

In the previous chapter, we have seen that convergent sequences are bounded. So there is an   with   for all  .

Now we can replace   within the bound by the constant   and we have   . Now, we proceed as for the second summand. We use the factor rule with   and get some   with   for all  , which in turn implies  .

Conclusion

It remains to choose a suitable  . For any   we need to have   and   . This can be done by choosing  , which only depends on  , as   and   also only depend on  .

Now, for all   there is

 

The power rule Bearbeiten

Theorem (Limit theorems for powers)

Let   be a sequence converging to  . Let   be any natural number . Then, the sequence   also converges with  .

How to get to the proof? (Limit theorems for powers)

The power rule is a consequence of the product rule- For the sequence   we have:

 

Repeating this step   times yields:

 

Now, the above argument with dots is not accepted as a proper proof in mathematics. The "clean" formulation is via induction in  .

Proof (Limit theorems for powers)

The proof is by induction, using the product rule.

Theorem whose validity shall be proven for the  :

 

1. Base case:

 

1. inductive step:

2a. inductive hypothesis:

 

2b. induction theorem:

 

2b. proof of induction step:

 

Example (An example for the power rule)

We can show for all   that the sequence   converges to  :

 

The quotient rule Bearbeiten

Theorem (Quotient rule for limits)

Let   be a sequence converging to   and   a sequence converging to   where   for all  . Then, the quotient sequence   converges with  .

How to get to the proof? (Quotient rule for limits)

It suffices to show   . Then, by the product rule, we get

 

In order to show  , we need to show that   gets arbitrarily small. We may use that   gets arbitrarily small, as   converges to   . So we need to re-formulate   in a suitable way:

 

The term   can be controlled, meaning we can make it arbitrarily small. The   in the denominator is just a constant and will not affect the convergence. The term   needs some more work, since it depends on  . More precisely, we need to bound   from above by a constant. That means, we require a lower bound for  .

By assumption,  , so   keeps a positive distance from 0. Due to convergence, we have an  , such that all sequence elements   with   satisfy   , i.e. they are not further away from   than half the distance from   to 0. Hence,   for all  . For the entire expression, that means

 

This expression can be made arbitrarily small, as   can be chosen arbitrarily small. For any given   we choose an   such that for all   there is

 

For all   it then follows that:

 

and we will be done with the proof. Now, let's write down the proof for   in a formal way.

Proof (Quotient rule for limits)

Let   be given. Since   there is an  , such that   for all   . In addition, there is an   with   for all  . Then, for  , we have:

 

Therefore,  . The product rule now implies

 

The root rule Bearbeiten

The following rule is in fact a generalization to the power rule above. It extends its validity from integer powers (like  ) to powers of the form  . Combining both rules, we get a limit theorem for all positive rational powers (like  ).

Theorem (Limit theorems for roots)

Let   be a non-negative sequence converging to  . In addition, let  . Then, the sequence   converges to  .

How to get to the proof? (Limit theorems for roots)

We need to find an upper bound for the absolute value   . What we can use is that   can be made arbitrarily small. Both expressions can be related by an auxiliary formula: For   and   there is

 

The reason why this holds is that the root function   is concave (meaning it curves down). You may draw this function yourself and verify why this holds. Taking the absolute of this equation, we get:

 

For  , we get

 

Hence,  . We apply this auxiliary formula with   and  :

 

The expression   can be made arbitrarily small by setting   "small enough":

 

So   is "small enough" and we have proven the desired inequality  .

Proof (Limit theorems for roots)

Let   and   be a non-negative sequence converging to  . In addition, let   be given. Since   there is an   with   for all  . Hence, for   there is

 

The monotony rule Bearbeiten

Theorem (Monotony rule for limits)

Let   and   be sequences converging to   and  . In addition, let   for almost all  . Then, we also have  .

Summary of proof (Monotony rule for limits)

The proof is by contradiction: We assume   and show that this cannot hold true.

Proof (Monotony rule for limits)

Assume  . Since   and  , for   (an interval width, which is half of the spacing between   and  ), there are thresholds   such that   for all   and   for all  . Hence for   (beyond both thresholds):

 

Therefore,   for all  . This contradicts   which must hold for almost all  . So the assumption   was wrong and   must hold.

Remarks concerning the monotony rule Bearbeiten

There is a special case: we consider a constant   :

Let   be a sequence covering to   and   (or analogously,  ) for almost all  . Then   (or analogously,  ).

The above proposition implies:

Let   be a convergent sequence and almost all elements are within the interval  . Then also the limit must be within the interval  .

Both cases can be connected: „ “ and „ “:

Let   and   be sequences converging to   and  , and let   for almost all  . Then, both limits also coincide:  .

Warning

The monotony rule does not hold for "  and  ", since within the limit, strict inequalities "become   and  ". An example are the sequences   and  . In this case   for all  , but in the limit, there is   . So the limit of   is not strictly smaller than  . Analogously, for all   there is   but in the limit, .