Prove that sequence $a_{n+1} = \frac{1}{1+a_n}$ is bounded? Consider sequence; 
$$a_1 = 1$$
$$a_{n+1} = \frac{1}{1+a_n}$$
Prove that sequence is bounded.
It is a question from my test. I couldn't figure out how to solve it.
 A: $$a_n > 0$$
$$a_{n+2}=\frac{1}{1+a_{n+1}}=\frac{1}{1+\frac{1}{1+a_n}}=\frac{1+a_n}{2+a_n}=1-\frac{1}{2+a_n}$$
$$\therefore \frac{1}{2}\leq a_{n}\leq 1\quad$$
A: First of all, you have to "feel" how does this sequence evolve. Computing the first terms, we have:
$$1, 0.5, 0.66666666, 0.6, 0.625, 0.615, 0.619, ....$$
We see that this sequence go up and down in a decreasing interval and that the values are all in $[0.5, 1]$. 
We have a feeling about what's happening. So now, you only have to prove (by recurrence) that for all $n\in \mathbb N$, $\frac{1}{2} \le a_n \le 1$. Do you follow that?
A: If $a_n$ is positive then
$$a_{n+1} = \frac{1}{1 + a_n} > 0.$$
Also, if $a_n$ is positive, then
$$1 < 1 + a_n \implies 1 > \frac{1}{1 + a_n} \implies 1 > a_{n+1}.$$
Therefore, if $a_{n} > 0$ then
$$0 < a_{n + 1} < 1.$$
That is, if all the elements of the sequence are positive then the sequence is bounded. Since the first element is positive ($a_1 = 1$), the first equation proves by induction that all the elements will be positive.
