Convergence of a recurrent sequence

Question

Sorry if this was already proved (that sequence converges to $\pi$) but I cannot find that proof. So recurrent sequence is : $$a_{n+1}=a_{n}+\sin(a_n)$$ and it seems if $0 < a_0 < 2\pi$ it converges to $\pi$.

If $\sin$ is replaced with $\cos$ it converges to – I think – $\frac{\pi}{2}$ in range $[0,3]$ at least

example Python program:

num=0.1
for i in range(10) :
    num+=math.sin(num)

result is : $3.141592653589793$

Again sorry if this is some well known fact - I'm not professional mathematician math is just hobby for me.

Answer 1

Just focusing on $a_{n+1}=a_n+\sin{(a_n)}$ as @Grey suggested $b_n=a_n-\pi$ $$a_{n+1}=a_n+\sin{(a_n)}\iff a_{n+1}-\pi=a_n-\pi-\sin{(a_n-\pi)}\iff \\ b_{n+1}=b_n-\sin{(b_n)}$$ and $b_0\in (-\pi,\pi)$.

Let's look at $f(x)=x-\sin{x}$ on $[-\pi,\pi]$ and show a few properties. Obviously $b_{n+1}=f(b_n)$

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Proposition 1. $f(x)$ is ascending.

Obviouly, since $f'(x)=1-\cos{x}\geq0$

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Proposition 2. If $b_0\in[0,\pi]$ then $0\leq f(b_0)\leq b_0$.

Because $0\leq b_0-\sin(b_0)\leq b_0$, as $\sin{x}\geq 0$ on $[0,\pi]$ and $0\leq \sin{x}\leq x$.

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Proposition 3. If $b_0\in[0,\pi]$ then $(b_n)_{n\in\mathbb{N}}$ is descending and bounded.

From Proposition 1 and 2

$$0\leq f(b_0)\leq b_0 \Rightarrow 0=f(0)\leq f(f(b_0))\leq f(b_0) \Rightarrow\\ 0\leq b_2\leq b_1 \leq b_0$$ and by induction $0\leq b_{n+1}\leq b_{n}\leq b_0$

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Proposition 4. If $b_0\in[-\pi,0)$ then $0\geq f(b_0)\geq b_0$.

Because $0\geq b_0-\sin(b_0)\geq b_0$, as $\sin{x}\leq0$ on $[-\pi,0)$ and $0\geq \sin{x}\geq x$.

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Proposition 5. If $b_0\in[-\pi,0)$ then $(b_n)_{n\in\mathbb{N}}$ is ascending and bounded.

From Proposition 1 and 4

$$0\geq f(b_0)\geq b_0 \Rightarrow 0=f(0)\geq f(f(b_0))\geq f(b_0) \Rightarrow\\ 0\geq b_2\geq b_1 \geq b_0$$ and by induction $0\geq b_{n+1}\geq b_{n}\geq b_0$

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Finally, in both cases we have a monotonic and bounded sequence, thus the limit of $(b_n)_{n\in\mathbb{N}}$ exists. Of course for $b_0=\pi$ or $b_0=-\pi$ we obtain constant sequences.

This also means that the limit of $(a_n)_{n\in\mathbb{N}}$ exists and you can find it from $L=L+\sin{L}$.