TITLE: Proving that the sequence $a_0=1, a_{n+1}=\cos(a_n)$ converges
QUESTION [5 upvotes]: Consider the following recursively defined sequence:
$$a_0 = 1, a_{n+1}=\cos\left(a_n\right)$$
Having a look with Wolfram Alpha, it's fairly clear that this sequence converges to something in the neighborhood of $0.74$ or so.
However, I have no clue how to actually prove that this sequence is convergent. I've thought about proving that it is a Cauchy sequence - that is, for any given $\varepsilon>0$, there exists an $N\in\mathbb{N}$ such that $|a_n-a_m|<\varepsilon$ for any $n,m>N$. 
However, I have no clue how to approach to repeated application of the cosine function, especially since you don't know how many iterations of the cosine function there are between $n$ and $m$.
Looking at the Wolfram Alpha plot of the first 30 terms in the sequence also gave me the idea to separate the whole thing into two subsequences $a_{2n}$ and $a_{2n+1}$ and then prove that they are increasing and decreasing while being bounded, but again I had no idea how to proceed to due to the repeated iteration of the cosine function.
I'd also be interested in a closed form of the limit if there is one, but I'd guess that none exists.

REPLY [6 votes]: By induction, it is easy to prove that
$\forall n\geq 0 \;\; 0\leq a_n \leq 1$.
let $f:x\mapsto \cos(x)$.
$f$ is decreasing from $[0,1]$ to $[0,1]$.
thus $(a_{2n})$ and $(a_{2n+1})$ are monotonic. on the other hand, by MVT,
$$|a_{ 2n+1 }-a_{2n}|=|a_{2n}-a_{2n-1}||\sin(\xi)|\leq \sin(1)|a_{2n}-a_{2n-1}|$$
This is just a Hint, you can finish it.