This is question 24 of a Brazilian Olympiad multiple-choice test (for undergraduate students) that you need to pass in order to qualify for the main national math Olympiad. The problem is as follows: Let $\{a_n\}$ be a sequence of real numbers recursively defined by $a_1=1$ and $a_{n+1}=\sin(a_n)$. Find the value of $$\lim_{n\to\infty}\frac{\log(a_n)}{\log(n)}.$$ The answer choices are
A) 0
B) $-\frac{1}{2}$
C) $-1$
D) 1
E) $-\infty$.
I tried using $\sin(x) > x – \frac{x^3}{6}$ to simplify the nested sine functions, but that didn't work. I also attempted to find the Taylor expansion of the nested sine function, but I could only get up to the $x^3$ term and the other terms were too difficult. My feeling is that since $a_n \to 0$ and $\log$ approaches $-\infty$ quickly while $\log(n)$ approaches $\infty$ very slowly, the answer will be E.
It's worth noting that this exam has 25 problems to be completed in 3 hours, and while there are many non-trivial problems, they shouldn't be too difficult.
Best Answer
It is easy to see that $a_n\to0$ as $n\to\infty$ and thus $$\frac{a_{n+1}-a_n}{a_n}=\frac{\sin a_n-a_n}{a_n}\sim -\frac16a_n^2, \qquad n\to\infty.$$
By Stolz-Cesàro theorem, we have \begin{align*} \lim_{n\to\infty}\frac{\log a_n}{\log n}&=\lim_{n\to\infty}\frac{\log a_{n+1}-\log a_n}{\log(n+1)-\log n}=\lim_{n\to\infty}n\log\left(1+\frac{a_{n+1}-a_n}{a_n}\right)\\ &=-\frac16\lim_{n\to\infty}na_n^2. \end{align*} So, we only need to compute the limit $\lim_{n\to\infty}na_n^2$. Using again Stolz-Cesàro theorem we obtain \begin{align*} \lim_{n\to\infty}na_n^2&=\lim_{n\to\infty}\frac n{\frac1{a_n^2}}=\lim_{n\to\infty}\frac{n+1-n}{\frac1{a_{n+1}^2}-\frac1{a_n^2}}=\lim_{x\to0}\frac1{\frac1{\sin^2x}-\frac1{x^2}}\\ &=\lim_{x\to0}\frac{x^2\sin^2x}{(x-\sin x)(x+\sin x)}=\lim_{x\to0}\frac{x^4}{\frac16x^3\cdot (2x)}=3. \end{align*} Therefore, $$\lim_{n\to\infty}\frac{\log a_n}{\log n}=-\frac16\times 3=-\frac12.$$