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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.

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  • $\begingroup$ $\lim\limits_{k\to\infty} a_k$ exists. Try to find out what it is. $\endgroup$ – Daniel Fischer Jul 26 '14 at 18:34
  • $\begingroup$ Or even easier, assume that it exists, and go from there. $\endgroup$ – Andrés E. Caicedo Jul 26 '14 at 21:43
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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:

$$\sum_{n=1}^N a_n \geq N x_0 - (1-x_0) = (N+1)x_0 -1.$$

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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?

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  • $\begingroup$ Hopefully no intuitionist will read my answer. $\endgroup$ – Git Gud Jul 26 '14 at 18:41
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    $\begingroup$ Either one does, or not. $\endgroup$ – Daniel Fischer Jul 26 '14 at 18:42

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