The sequence $a_{0}, a_{1}, \ldots$ is defined by the recursive formula $a_{n+1}=\sin a_{n}$ starting from an arbitrary value of $a_{0}$. Is this sequence convergent, and if so, what is its limit?
Since the range of the $\sin x$ function is the closed interval $[-1,1]$, $\left|a_{1}\right|=\left|\sin a_{0}\right| \leq 1$. Let us denote $\left|a_{1}\right|$ by $x_{1}$, and starting from $x_{1}$, define the terms of the sequence by the recursion

$$
x_{n+1}=\sin x_{n}, \quad n=1,2, \ldots
$$

If $a_{1} \geq 0$, then the sequences $a_{n}$ and $x_{n}$ are identical. If $a_{1}<0$, then the sequences $a_{n}$ and $-x_{n}$ are identical. Therefore, it is sufficient to consider the $x_{n}$ sequence. We will show that this sequence converges to 0. Suppose, for the sake of contradiction, that the sequence does not converge to 0. Then there exists a positive number $\varepsilon$ such that for any natural number $n$, there is an index $N \geq n$ for which

$$
\left|x_{N}-0\right| \geq \varepsilon
$$

According to (2), $x_{N}>0$, so by (3), $x_{N} \geq \varepsilon$. Since the sequence is monotonically decreasing, this implies that $x_{k} \geq \varepsilon$ for all $k \leq N$, and in particular for all $k \leq n$. Since $n$ is arbitrary, we have

$$
x_{n} \geq \varepsilon, \quad n=1,2, \ldots
$$

We will show that the function $f(x)=x-\sin x$ is positive for $0<x<1$. Since $f(0)=0$ and $f^{\prime}(x)=1-\cos x>0$ for $0<x<1$, $f(x)$ is strictly increasing on this interval, and thus $f(x)>0$ for $0<x<1$.

Therefore, in case (4),

$$
x_{n}-x_{n+1}=f\left(x_{n}\right)>f(\varepsilon)>0
$$

holds for any $n$, which implies

$$
x_{1}>x_{1}-x_{n+1}=\left(x_{1}-x_{2}\right)+\left(x_{2}-x_{3}\right)+\ldots+\left(x_{n}-x_{n+1}\right)>n f(\varepsilon)
$$

This is impossible, since $f(\varepsilon)>0$ and the right-hand side can be arbitrarily large, while $x_{1}$ is a fixed number.

This shows that the assumption that the sequence $x_{n}$ does not converge to 0 leads to a contradiction, so $x_{n}$ is convergent, and its limit is 0.

Remark. In our solution, the indirect assumption was not that $x_{n}$ is convergent but its limit is not 0, but rather that it is not true that $x_{n}$ is convergent and its limit is 0. This means that $x_{n}$ is either not convergent, or it is convergent but its limit is not 0. Generally, if a sequence is monotonically decreasing and bounded below, then it is convergent - however, we did not want to use this theorem in our solution, as it is not part of the high school curriculum.[^0]

[^0]:    ${ }^{1}$ See the solution of F 1989 in this issue.