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I know that this sequence is bounded and also that if it converges, then it should converge at $x$ such that $x = \cos x$. Assuming I am not allowed to use calculator (and hence not able to solve $x = \cos x$ numerically), how would I prove the convergence?

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marked as duplicate by rtybase, Namaste, Shailesh, GNUSupporter 8964民主女神 地下教會, Brian Borchers Jan 29 '18 at 3:23

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Note that for all $n \in \mathbb{N}$ $$ x_n \leq 1 $$ because $\cos(x) \leq 1$. Then you can use study the function $f : x \mapsto \cos\left(x\right)-x$. $$ f'\left(x\right)=-\sin\left(x\right)-1 $$ The funcion is decreasing and $f(0)=1$ and $\displaystyle f\left(\frac{\pi}{2}\right)=0$. Hence the interval $\displaystyle \left[0,\frac{\pi}{2}\right]$ is the only interval where $f$ is positive on $\mathbb{R}^{+}$. Hence it would be simpler for you to show that the sequence $\left(x_{2n}\right)_{n \in \mathbb{N}}$ and $\left(x_{2n+1}\right)_{n \in \mathbb{N}}$ are adjacent sequences. You will show that those two sequences converge toward a same limit that satisfies $$ \cos\left(\ell\right)=\ell $$ which you should know, has only one solution on $\displaystyle \left[0,\frac{\pi}{2}\right]$.

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The sequence is bounded in $[-1,1]$ (since $\lvert \cos\rvert \leq 1$) and of the form $$ x_{n+1} = f(x_n) \tag{1} $$ with $\lvert f'\rvert = \lvert \sin\rvert <1$ on $[-1,1]$. It follows that $f$ is Lipschitz on $[-1,1]$ with Lipschitz constant $\sup_{[-1,1]}\lvert \sin\rvert<1$, and therefore the sequence converges by fixed-point theorems to the (unique) solution of $$x=\cos x\tag{2}$$ in $[-1,1]$.

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