Limit proofs using the -definition are quite laborious. In this chapter we will study some limit theorems that simplifies matters.
The limit theorems are the following:
Theorem (Limit theorems)
Let and be two convergent sequences with and . Let also be arbitrary. Then we have
- for all
If furthermore and for all , then we also have
For and for all :
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 :
Let be a sequence. If , then converges to zero, i.e. .
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.
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,
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 .
There is so we have convergence
Choose such that for all . Such an can be found, since converges to . Then,
this shown the assumption .
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 .
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
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:
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 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
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.
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: .
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,.