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Let x ∈ R.

Consider the sequence ${a_n}$, where $a_1$ = x and $ a_{n+1 }= cos(a_n)$.

From picture I can observe that ${a_n}$ converges. But how to prove it analytically. I tried to show it is Cauchy. But that doesn't work.

Because $| a_n-a_m| \leq |a_n- cos a_{n+1}|+....+|a_{m-1}- a_m|$.

But I don't know how to prove $ |a_n- cos a_{n+1}|$ approaches zero. Any help (other methods too) will be appreciated.

Edit: There is a question which deals exactly with the same sequence. But That question doesn't have the proof for the convergence of the sequence. The answers for that question just state that the sequence is convergent. But I want to know how it's convergent. That question actually deals with the denseness and not with convergence

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  • $\begingroup$ @Robert Z That question stated above doesn't have the proof for the convergence of the sequence. The answers for that questionjust state that the sequence is convergent. But I want to know how it's convergent. That question actually deals with the denseness and not with convergence $\endgroup$ – Cloud JR Oct 20 '18 at 17:50
  • $\begingroup$ " (which is also a contraction because of MVT), this function has a fixed (stationary) point. So, regardless of $a_0$, $a_1$ will end in between $[−1,1]$ and from there on, the sequence $\{a_n\}$ will tend to the fixed point of $\cos(x)$." This is more than enough to complete the proof yourself. Plus that link is linked to other similar questions. $\endgroup$ – rtybase Oct 20 '18 at 18:08
  • $\begingroup$ More of the duplicates math.stackexchange.com/questions/2071411/… $\endgroup$ – rtybase Oct 20 '18 at 19:25
  • $\begingroup$ And another one math.stackexchange.com/questions/2144691/… $\endgroup$ – rtybase Oct 20 '18 at 19:26
  • $\begingroup$ And another one math.stackexchange.com/questions/2382210/… $\endgroup$ – rtybase Oct 20 '18 at 19:27

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