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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.

Tom
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    Look for the method of fixed point interations. There is a quite general result on convergence of the method which works in your case. – Olod Feb 14 '17 at 21:49
  • Excuse me, Sir! Can you show me some magic right now? – SAJW Feb 14 '17 at 21:55

2 Answers2

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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.

hamam_Abdallah
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Draw a graph of $y=x$ and $y= \cos x$. Any possible limit has $y=\cos y$ so the points of intersection of the curves are possible limts. Then start bpouncing from one curve to the other: given a $\cos a_n$ go horizontally to hit the linear graph, then vertically to hit the $\cos$ graph and so on. If this does not make immediate sense look up pictures of how Newton Raphson convereges (or not) for squre roots.

mike stone
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