I want to show here how to solve the following problem from this week’s homework set:

Starting with a given , define the subsequent terms of a sequence by setting . Determine whether the sequence converges, and if it does, find its limit.

This is a nice simple example of a (discrete) dynamical system in one variable. It turns out that the sequence always converges, but the limit depends on the value of

Case 1. for some

In this case , so the sequence trivially converges.

Case 2. for some

In this case I will show that and that

First, notice that if , then we can write for some with and , so

Second, recall that for all . You are probably familiar with this inequality from Calculus I; if not, one can prove it easily as follows: Let , so . Also, for all , so is always decreasing, and the result follows.

Also, recall that , so

We have shown (by induction) that the sequence is strictly increasing and bounded above (by ). Thus, it converges. If is its limit, then

so and since , it follows that

Case 3. for some

In this case, and

The argument is very similar to the one for Case 2: If , then so , and for some , so and therefore . It follows that the sequence is decreasing and bounded below (by ), so it converges. As before, the limit must in fact be , and we are done.

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