Prove limit of sequence

Question

Assume $\space a_0=1 $ and for each natural n, $a_n = sin(a_{n-1}) .$ prove: $$ \lim\limits_{n\to \infty} n a_{n} ^2=\frac{1}{3} $$

Answer 1

Clearly $\lim_{n\to\infty}a_n=0$. By Stolz-Cesaro's Theorem \begin{eqnarray} \lim\limits_{n\to \infty} n a_{n} ^2&=&\lim\limits_{n\to \infty} \frac{n}{\frac1{a_{n}^2}}=\lim\limits_{n\to \infty} \frac{1}{\frac1{a_{n+1}^2}-\frac1{a_{n}^2}}=\lim\limits_{n\to \infty} \frac{a_{n}^2a_{n+1}^2}{a_{n}^2-a_{n+1}^2}=\lim\limits_{n\to\infty}\frac{a_n^2\sin^2a_n}{a_n^2-\sin^2a_n}\\ &=&\lim\limits_{x\to0}\frac{x^2\sin^2x}{x^2-\sin^2x}=\lim\limits_{x\to0}\frac{\sin^2x}{x^2}\frac{x^4}{x^2-\sin^2x}=3. \end{eqnarray}