what are the properties of this sequence?

Question

Can we assert that this sequence is convergent or divergent?

$$u_o = a, \text{ and } \forall n \in \mathbb{N} : u_{n+1} = \sin(u_n)$$

What is the limit of $u_{n}$ if a=$\frac{\pi}{4}$

Answer 1

for any $a>0$: $$0\le u_1=\sin(a)\le 1$$ for any $x \in [0;1]$: $\sin(x)<x$ thats why sequence is decreasing and all elements of this sequence is positive. Thats why the sequence have limit.

Answer 2

The limit shown by aid78 corresponds to a fixed point hence satisfies $$ \sin(x)=x $$ which gives us one possible limit : $0$.

So $\displaystyle \left(u_n\right)_{n \in \mathbb{N}}$ converges to $0$.

Answer 3

It is convergent, and the limit is zero. It follows quite nicely from interpreting the sine geometrically. Given $a\in \mathbb{R}\setminus \lbrace 0\rbrace$, then $\mathrm{sin}(a)\in [-1,1]$ is the height given by the arc of $a$ on the unit circle. At most $\mathrm{sin}(a)=\pm 1$ and subsequent sines are shorter in length since the arc $\mathrm{sin}(\pi /2)> \mathrm{sin}(1) > \cdots$ (think about it), i.e. $$|u_1| > |u_2| > |u_3| > \dots $$ So this tends to zero. Note that for $a=0$, this would yield $|u_i|=|u_j|=0$ for all $i,j\in \mathbb{N}$.

Answer 4

Fast answer. This sequence tends to zero, relatively slowly, in a similar fashion to the sequence $$ |b_0|<1, \quad b_{n+1}=b_n-\frac{b_n^3}{6}. $$

Explanation. Clearly, $u_n\to 0$, since it is strictly decreasing, if $\sin u_1>0$ or strictly increasing, if $\sin u_1<0$, due to the fact that $0<\sin x<x$, if $\frac{\pi}{2}>x>0$, or $x<\sin x<0$, if $-\frac{\pi}{2}<x<0$,

Using Stoltz Theorem, we get that $$ \lim\frac{u_{n}}{n^k}=\lim\frac{u_{n+1}-u_n}{kn^{k-1}}=-\lim\frac{u_n-\sin{u_n}}{kn^{k-1}}=-\lim\frac{u_n^3}{6kn^{k-1}} $$ Hence $$ k-1=3k\quad\Longrightarrow\quad k=-\frac{1}{2} $$ and therefore $u_n=\mathcal O(n^{-1/2})$.

Next, we obtain that $$ a=\lim\frac{u_{n}}{n^{-1/2}}=\lim\frac{u_n^3}{3n^{-3/2}}=\frac{a^3}{3} $$ and thus $$ \frac{u_n}{\sqrt{3}n^{-1/2}}\to 1. $$