# Verifying a sequence that converges to pi [closed]

A computer program ouputs the digits of $\pi$ by evaluating the recurrence relation

$a_{n+1} = a_n + sin \ a_n$

with $a_0 = \frac{6}{5}$

Does the sequence actually converge or is this just coincidence?

I would like to answer this by rewriting it as a differential equation.

Can I write $\frac{dy}{dx} = sin \ x$ and solve this?

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## closed as too localized by Gjergji Zaimi, Harald Hanche-Olsen, Pietro Majer, Todd Trimble♦, Gerry MyersonDec 26 '10 at 15:34

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The function $f(a):=a+\sin(a)$ maps the interval $[0,\pi]$ into itself, keeping fixed the endpoints. Moreover $f(a) > a$ for $0 < a < \pi$. So any starting point $0 < a_0 \le \pi$ generates an increasing bounded sequence; hence it converges; by continuity of $f$ the limit is a fixed point; hence it's $\pi$. Since $f'(\pi)=0$ and $f''(\pi)=0$ the convergence is cubic; precisely $0 < \pi-a_ {n+1} \le (\pi- a_n)^3 /6$. –  Pietro Majer Dec 26 '10 at 14:25
So it seems the initial value doesn't matter just as long as its in $(0,\pi]$. –  user11822 Dec 26 '10 at 19:21

But you can write $\Delta a_x=\sin a_x$.
Because $\sin(\pi) = 0$, $a_n = \pi$ is a fixed point.