Convergence/Divergence of a the series $\sum_{k=1}^{\infty} a_k$, where $a_1=1$ and $\forall 1\leq k\in\mathbb{N},a_{k+1}=\cos(a_k)$ I got this question:
Determine wether the series $\sum_{k=1}^{\infty} a_k$ absolutely converges, conditionally converges or diverges, where $a_1=1$ and for each $1\leq k \in\mathbb{N}$, $a_{k+1}=\cos(a_k)$.
I think this series diverges but I tried for an hour and half and wasn't able to proceed.
I tried to show that $lim_{k\to\infty}a_k\neq 0$ By showing that $lim_{k\to\infty}a_k$ does not exist by the definition of the limit of sequence and so by the divergence test the series diverges but I failed.
Any help will be appreciated.
 A: Following up (sort of) on Daniel Fischer's hint, the limit $\lim \limits_{n\to \infty}\left(a_n\right)$ either exists or doesn't exist.
If it doesn't exist, $\boxed{\_\_\_\_}$.
If it exists and it equals $l$, you have $l=\cos(l)$. For the series to converge, what would $l$ need to be?
A: Since the function $f(x)=\cos x$ is decreasing and concave on the $[0,1]$ interval, in such an interval there is only one root of $x-\cos(x)$ - namely, around 
$$x_0=0.73908513321516\ldots\approx\frac{17}{23}$$ and the sequence $\{a_n\}_{n\in\mathbb{N}_0}$ given by:
$$ a_1=1, a_{n+1}=\cos(a_n) $$
converges towards $x_0$, with alternating signs for the error term $a_n-x_0$. Moreover,
$$|a_{n+1}-x_0|< |a_n - x_0|= O\left((\varepsilon+\sin x_0)^n\right) \ll \left(\frac{7}{10}\right)^n, $$
since the cosine is a Lipschitz function, and the asymptotics for the error term follows from considering the Taylor series of the cosine function in a neighbourhood of $x_0$.
Now, since $x_0=\cos(x_0)>0$, the series $\sum_{n=1}^{+\infty} a_n$ diverges to $+\infty$. 
With even more accuracy, since the error term has an alternating sign:
$$\sum_{n=1}^N a_n \geq N x_0 - (1-x_0) = (N+1)x_0 -1.$$
