Is the recurrent sequence $a_{n+1}=\arcsin (a_n)$ converging? A recurrent sequence $a_{n+1}=\arcsin (a_n)$, $a_1=a$ $(a\in [-1,1])$ is given. If, for example, $a=1$, then $a_3$ does not exist. But does the sequence exist and converge if $a$ is a small enough positive number?
My Attempt (copied from answer)
Given sequence is bounded because range of function $y=\arcsin(x)$ is $[-\pi/2,\pi/2]$.
This sequence is strictly increasing when $a_1\gt0$ and strictly decreasing when $a_1\lt0$ (for example, because for any $t$ from $(0,\pi/2)$, $\sin(t)$ is less than $t$ and so if $a_1\gt0$,  $a_{n+1}\gt \sin(a_{n+1})=a_n$ for any $n$).
For $a_1=0$ this sequence is constant.
So the sequence is monotonic.
So this sequence, being bounded and monotonic, converges for any $a_1$.
 A: $\{a\}_{i=1}^\infty$ is converging for only $a_1=0$. If $a_1=0$, $a_1$ is a fixed point of $\arcsin$, so we are done.
If $a_1>0$, then the sequence is increasing as $\arcsin(x)\geq x$ for positive $x$. $\arcsin(x)>x+\frac{x^3}{6}$ (Maclaurin Series), as $a_n$ is increasing, we have
$$a_{1+\left\lceil\frac{6}{a_1^3}\right\rceil}>a_1+\left\lceil\frac{6}{a_1^3}\right\rceil\frac{a_1^3}{6}>1$$
If $a_1<0$, then the sequence is decreasing as $\arcsin(x)\leq x$ for negative $x$. $\arcsin(x)<x+\frac{x^3}{6}$ (Maclaurin Series), as $a_n$ is decreasing, we have
$$a_{1+\left\lceil\frac{6}{|a_1|^3}\right\rceil}>a_1+\left\lceil\frac{6}{|a_1|^3}\right\rceil\frac{a_1^3}{6}>-1$$
So $a_n$ is eventually not a real number, hence the sequence is not converging in this case.
A: Assuming $a_1\gt0$,
$$
\begin{align}
a_{n+1}
&=\arcsin(a_n)\\
&=\int_0^{a_n}\frac{\mathrm{d}t}{\sqrt{1-t^2}}\\
&\ge\int_0^{a_n}\left(1+\frac{t^2}2\right)\mathrm{d}t\\
&=a_n+\frac{a_n^3}6\\
&=a_n\left(1+\frac{a_n^2}6\right)
\end{align}
$$
Thus,
$$
\begin{align}
a_n
&\ge a_1\prod_{k=1}^{n-1}\left(1+\frac{a_k^2}6\right)\\
&\ge a_1\left(1+\frac{a_1^2}6\right)^{n-1}\\
\end{align}
$$
and therefore, for some $n$ no greater than
$$
\frac{\log\left(\frac1{a_1}\right)}{\log\left(1+\frac{a_1^2}6\right)}+2
$$
we have $a_n\gt1$.
Thus, for any $a_1\gt0$, there is an $n$ so that $a_n\gt1$. Since $\sin(-x)=-\sin(x)$, for any $a_1\lt0$, there is an $n$ so that $a_n\lt-1$.
Thus, only for $a_1=0$ does the sequence continue without reaching a value outside of $[-1,1]$.
A: The sequence is convergent only for $a_1=0$ because $\arcsin(x)$ for $x>0$ is convex and $\arcsin(x)>x$, and for $x<0$ concave and $\arcsin(x)<x$.
