What does recursive cosine sequence converge to? I have a sequence defined as follow:
$a_0 = 1, a_n=\cos\left(a_{n-1}\right)$.
I want to count $\lim_{n\rightarrow\infty} a_n$ - it definitely does have limit by looking at the graph, the first few numbers of the limit are 0.7390851, but I have no idea, if that number is related to some other real number ($\pi$ or something like that).
The sequence is from this site, but they don't provide actual result for their own sequence. 
 A: This is a standard trick worth knowing.
Supposing the limit does exist, call it $x$. If $x$ is the limit of the sequence, it has the property that $x = \mbox{cos}(x)$. From a graph, we can see that there is exactly one solution. Lastly, Wolfram Alpha tells us that $x = \mbox{cos}(x)$ has the solution $x = 0.739085$ as you said.
A: This happens to be a relatively well-known number. It is called the Dottie Number, named after the not-at-all famous Professor of French, Dottie. I should also point out that the number is transcendental (also pointed out in the link).
A: Here is an elementary proof of convergence of the sequence:
Notice that $0 \leq a_n \leq 1$ for all $n$.
Consider the function $f(x) = x - \cos \cos x$.
This is increasing in $[0,1]$.
Now since $f(0) \lt 0$ and $f(1) \gt 0$, $f(x) = 0$ has a unique root (say $D$) in $(0,1)$, which is also the root of $x = \cos x$.
Now if $a_n \lt D$, then $a_n - a_{n+2} = f(a_n) \lt 0$
if $a_n \gt D$, then $a_n - a_{n+2} = f(a_n) \gt 0$
We also have that $g(x) = \cos x - D$ is decreasing in $[0,1]$ and thus if $a_n \lt D$ then $a_{n+1} \gt D$ and if $a_n \gt D$, then $a_{n+1} \lt D$.
Since $a_0 = 1 \gt D$
The sub-sequence $a_0, a_2, a_4, \dots$ is monotonically decreasing and bounded below and hence is convergent (to $D$).
Similarly, the sub-sequence $a_1, a_3, a_5, \dots$ is monotonically increasing and bounded above, and is convergent (to $D$).
Thus $\lim a_n = D$
A: The limit is the (unique) solution of the equation $\cos x = x$. This can't be expressed in any simpler way.
