Recurrence that converge to $\pi/2$

Question

I've saw this kind of the question and I can't find the way to prove.

I want to know how the step of proving $x_n = x_{n-1} + \cos (x_{n-1} ) $ converges to $\frac{\pi}2$.

The $x_1$ equals to 1.

Answer 1

You cannot prove it because the sequence is not defined. It is not known what $x_1$ is equal to, and therefore, we don't know what $x_2$ is, and what $x_3$ is, and ...

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Furthermore, if $x_1=-\frac\pi 2$, then the sequence does not converge to $\frac\pi2$ because it converges to $-\frac\pi2$.

Answer 2

The limit point depends on the initial term $x_0$:

i) if $x_0\in I_k=(\pi/2+(k-1)\pi,\pi/2+k\pi]$ with $k$ an even integer then the sequence remains in $I_k$, it is increasing (because $\cos(x)>0$ inside $I_k$) and $x_n\to \pi/2+k\pi$;

ii) if $x_0\in J_k=[\pi/2+(k-1)\pi,\pi/2+k\pi)$ with $k$ an odd integer then the sequence remains in $J_k$, it is decreasing (because $\cos(x)<0$ inside $J_k$) and $x_n\to \pi/2+(k-1)\pi$.