IMC 2010 · Day 2 · P6

(a) NO. For example, for $x_1 = \pi$ we have $x_n = (-1)^{n-1} \pi$, and the sequence is divergent.

(b) YES. Notice that $|y_n|$ is nonincreasing and hence converges to some number $a \ge 0$.

If $a = 0$, then $\lim y_n = 0$ and we are done. If $a > 0$, then $a = \lim |y_{n+1}| = \lim |y_n \sin y_n| = a \cdot |\sin a|$, so $\sin a = \pm 1$ and $a = (k + \frac{1}{2}) \pi$ for some nonnegative integer $k$.

Since the sequence $|y_n|$ is nonincreasing, there exists an index $n_0$ such that $(k + \frac{1}{2}) \pi \le |y_n| < (k+1) \pi$ for all $n > n_0$. Then all the numbers $y_{n_0+1}, y_{n_0+2}, \dots$ lie in the union of the intervals $\left[ (k + \frac{1}{2}) \pi, (k+1) \pi \right)$ and $\left( -(k+1) \pi, -(k + \frac{1}{2}) \pi \right]$.

Depending on the parity of $k$, in one of the intervals $\left[ (k + \frac{1}{2}) \pi, (k+1) \pi \right)$ and $\left( -(k+1) \pi, -(k + \frac{1}{2}) \pi \right]$ the values of the sine function is positive; denote this interval by $I_+$. In the other interval the sine function is negative; denote this interval by $I_-$. If $y_n \in I_-$ for some $n > n_0$ then $y_n$ and $y_{n+1} = y_n \sin y_n$ have opposite signs, so $y_{n+1} \in I_+$. On the other hand, if If $y_n \in I_+$ for some $n > n_0$ then $y_n$ and $y_{n+1}$ have the same sign, so $y_{n+1} \in I_+$. In both cases, $y_{n+1} \in I_+$.

We obtained that the numbers $y_{n_0+2}, y_{n_0+3}, \dots$ lie in $I_+$, so they have the same sign. Since $|y_n|$ is convergent, this implies that the sequence $(y_n)$ is convergent as well.

(for part (b)) Similarly to the first solution, $|y_n| \to a$ for some real number $a$.

Notice that $t \cdot \sin t = (-t) \sin(-t) = |t| \sin |t|$ for all real $t$, hence $y_{n+1} = |y_n| \sin |y_n|$ for all $n \ge 2$. Since the function $t \mapsto t \sin t$ is continuous, $y_{n+1} = |y_n| \sin |y_n| \to |a| \sin |a| = a$.