Series with a recursive sequence? converges or diverges?

Question

$$\sum _{ n=1 }^{ \infty }{ { a }_{ n } } $$ where $${ a }_{ n }=sin({ a }_{ n-1 })$$ and $${ a }_{ 0 }=1$$ I know the limit of an is zero but how do I know if the series converges or diverges? Can I show it using one of the tests?

Answer 1

Let's first prove that $a_n \ge 1/(n+1)$ by induction. It's clearly true for $n=0$. Now assume it's true for some $k>0$, i.e., $1 \ge a_k \ge 1/(k+1)$. Since $\sin x$ is increasing on $[0,1]$, we have $a_{k+1}=\sin a_k \ge \sin \frac{1}{k+1}.$ And we know that $\sin x \ge x - \frac{1}{3!}x^3$ for positive $x$; so $$\sin\frac{1}{k+1}\ge \frac{1}{k+1}-\frac{1}{6(k+1)^3} > \frac{1}{k+1} - \frac{1}{(k+1)(k+2)}=\frac{1}{k+2}.$$ Therefore, $a_k \ge 1/(k+1) \implies a_{k+1} \ge 1/(k+2)$.

We conclude that $a_n \ge 1/(n+1)$ for all $n\ge 0$; and hence the series of iterated sines diverges by comparison with the harmonic series.