Nested sequence and series

Let $a_n =\underbrace{\sin \left ( \sin \cdots \left ( \sin x \right )\cdots \right )}_{n \;\; {\rm times}}$. Examine if the series $\sum \limits_{n=1}^{\infty} a_n$ converges.

Solution



We are proving that the series diverges. Let $s_n$ denote the $n$ nested $\sin x$. It is easy to see using the inequality $\sin x< x$ that $s_n \rightarrow 0$. We are proving that $s_n \approx \frac{c}{\sqrt{n-1}}, \; c>0$.

We have that:

$$\begin{aligned}
\frac{1}{s_n^2} &=\frac{1}{\sin^2 (s_{n-1})} \\
 &=\frac{1}{\left ( s_{n-1} - \frac{1}{3!}s_{n-1}^3 + \mathcal{O}\left ( s_{n-1}^5 \right ) \right )^2} \\
 &= \frac {1} {s_{n-1}^2 } \left (1 - \frac {1}{3!} s_{n-1} ^2+ \mathcal{O} \left (s_{n-1}^3\right )\right )^{-2} \\
 &= \frac {1} {s_{n-1}^2 } \left (1 + \frac {1}{3} s_{n-1} ^2+ \mathcal{O} \left (s_{n-1}^3\right )\right )\\
 &= \frac {1} {s_{n-1}^2 }  + \frac {1}{3} + \mathcal{O} \left (s_{n-1}\right ) 
\end{aligned}$$

Adding by parts we have that:

$$\frac{1}{s_n^2}= \frac{n-1}{3}+ \mathcal{O}\left ( \sum_{k=1}^{n-1}s_k \right )$$

and hence:

$$\frac{1}{(n-1)s_n^2} = \frac{1}{3}+ \frac{1}{n-1}\mathcal{O}\left ( \sum_{k=1}^{n-1} s_k \right )  \tag{1}$$

We note that as $n \rightarrow +\infty$ from Cesàro and the fact that $s_n \rightarrow 0$ we get $\displaystyle \frac{1}{n-1} \sum_{k=1}^{n-1}s_k \rightarrow 0$.

Taking limits back at $(1)$ we deduce that:

$$\frac{1}{(n-1)s_n^2} \rightarrow \frac{1}{3}\Rightarrow  s_n \approx \sqrt{\frac{3}{n-1}}$$

and the proof is complete.