This article shows you how to prove convergence for recursively defined sequences, i.e. if there is only given and it is not known, what explicitly looks like. The monotony criterion will turn out very useful for this.
Let be a sequence defined by and . Does it converge? If yes, what is its limit value?
Applying the epsilon-criterion is complicated, here: it is not a priori known how elements with large indices look like (e.g. ), which makes it hard to guess a limit . It is even harder to give a bound for , as the explicit form of is often not known.
The article will cover two strategies for proving convergence of a recursively defined sequence:
Finding the explicit form: You may try to find the explicit for of the sequence elements . Then it can be easier to determine a limit and prove convergence.
Using the monotony criterion: If you can not find an explicit form, but you are convinced that the sequence should converge, you may try to use this criterion. It works without knowing an explicit form.
One way to prove convergence is to try to find an explicit form for the sequence elements . It is useful, to compute the first sequence elements in order to get a feeling of how the sequence behaves. In the above example:
These elemnts are all smaller than 1, while slowly approaching it. We therefore assert that converges to 1. Let us try to find an explicit form, i.e. a map . That means, we are looking for a term with:
The denominator suspiciously looks like a power of . The enumerator seems to be always one less than the denominator:
That would mean, . So we assert that
Now, we should try to prove or disprove this assertion. Induction is a suitable way to do this, as the recursion relation may perfectly serve as an induction step. The induction starts with
The induction step from to is done using the recursion relation:
So we indeed proved . This explicit form allows us to use the usual tools (from the previous articles) for proving convergence. For instance, we know and we can use the limit theorems:
Alternative way: using a telescoping sum[Bearbeiten]
Sometimes, finding a simple explicit form may be enormously hard or even impossible. In any case, one can write the elements using telescoping sums. To illustrate this, we consider:
The difference between two sequence elements and is
Applying this step again yields
So each time, we get an additional factor of . After steps, there will be
This is not yet an explicit for for , but for the differences . However, we can express using telescoping sums:
What if we cannot find an explicit form of the sequence elements? If the sequence is monotonously increasing or decreasing, the monotony criterion can be useful. We recall this criterion:
All bounded and monotonous functions converge.
Hence, it suffices to show boundedness and monotony of the sequence. For instance, in the above case:
Which seems to be monotonously increasing and bounded by .
Monotony is proven inductively. Clearly , as
The induction step from to consists of showing that the sequence element gets bigger within that step:
This establishes monotony of the sequence. Boundedness by can also be shown inductively. Of course, which is smaller than 1. In the induction step, we need to show that if , then also stays smaller or equal 1:
Hence, is bounded from above by . Now, the monotony criterion can be applied: The sequence is monotonously increasing and bounded from above, so it must converge.
The convergence above can be used for finding the limit. By step 1, we already know that there must be a limit with and it can be at most the upper bound, namely 1. As the sequence elements approach 1 closer and closer, we assert that it is exactly 1.
To verify this, we take a look at :
So if there is any , it must fulfil . We resolve the equation for and get
So is indeed the limit of the sequence .
Question: Above, we used that if is known, then there is also . Why is this the case?
is the limit of the sequence . This is just a subsequence of , which contains all but the first element. Since each subsequence converges to the same limit as the original one, we have .
Exercise (Proving convergence of a recursively defined sequence)
We consider the following recursive sequence
Prove that this sequence converges and determine the limit.
What happens with , concerning convergence?
Solution (Proving convergence of a recursively defined sequence)
Solution sub-exercise 1:
Applying this step times, we get
We use telescoping sums
So we have an explicit form. With and the limit theorems, we get the limit
Solution sub-exercise 2:
For and we can plug in
The next plugging-ins would exactly look the same and we always get 1. So we assert that the sequence is constantly 1 and therefore converges to 1. We prove this assertion by induction
Theorem whose validity shall be proven for the :
1. Base case:
We start with . As above, and renders .
1. inductive step:
2a. inductive hypothesis:
2b. induction theorem:
2b. proof of induction step:
Hence, the sequence is constantly 1 and its limit is .
Application: computing roots by Heron's method[Bearbeiten]
We will now come to an application, where convergence of a recursively defined sequence can be shown using the monotony criterion: Heron's method can be used to compute roots of integers, rational or even real numbers . It recursively constructs a sequence of better and better approximations for the root . That these approximation get increasingly better is fixed within a theorem:
Theorem (Heron's method)
Let be a real number with . Let the sequence be recursively defined by
Then, converges to .
Sequence elements of yield an increasingly precise approximation for , where we can get to an arbitrary precision . One can, for instance, use a computer to approximate up to an arbitrary precision, which is done by performing a sufficiently large number of recursion steps.
Proof (Heron's method)
In order to show that really converges to , we will show that:
converges by the monotony criterion.
The limit of is .
We can not leave out the first step, as in step 2, we use that the sequence converges. A direct proof of convergence using the Epsilon-criterion (in one step) would be way more complicated.
Proof step: The sequence converges.
We prove that is monotonously decreasing and bounded from below.
Proof step: The sequence is bounded from below.
A bound is established using the inequality between arithmetic and geometric means: For there is . Hence,
So is bounded from below by .
Proof step: The sequence is monotonously decreasing.
This means, we need to show , or equivalently, :
So is bounded from below and monotonously decreasing. Hence, it converges by means of the monotony criterion.
Proof step: The limit of is .
We will make use of a trick, here: As converges, there must be a limit , i.e. a real number with . Then, there is also , since is a subsequence of . The recursion relation and the limit theorems yield:
This can be resolved for and we get the limit:
At this point, there are two candidates for the limit. But since , we also have so the convergence must be towards the positive limit candidate
The initial value can be chosen to be any (strictly) positive real number. The proof of convergence works exactly the same way.
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