Putnam 2020, Problem A-3

Let \( a_0 = \pi/2 \), and let \( a_n = \sin(a_{n-1}) \) for \( n \geq 1 \). Determine whether
\[
\sum_{n=1}^{\infty} a_n^2
\]
converges.

Solution:
The series diverges. First note that since $\sin(x) < x$ for all $x > 0$, the sequence $\{a_n\}$ is positive and decreasing, with $a_1 = 1$. Next, we observe that for $x \in [0,1]$, $\sin(x) \geq x - x^3/6$: this follows from Taylor's theorem with remainder, since $\sin(x) = x - x^3/6 + (\sin c) x^4/24$ for some $c$ between $0$ and $x$.

We now claim that $a_n \geq 1/\sqrt{n}$ for all $n \geq 1$; it follows that $\sum a_n^2$ diverges since $\sum 1/n$ diverges. To prove the claim, we induct on $n$, with $n = 1$ being trivial. Suppose that $a_n \geq 1/\sqrt{n}$. To prove $\sin(a_n) \geq 1/\sqrt{n+1}$, note that since $\sin(a_n) \geq \sin(1/\sqrt{n})$, it suffices to prove that $x - x^3/6 \geq (n+1)^{-1/2}$ where $x = 1/\sqrt{n}$. Squaring both sides and clearing denominators, we find that this is equivalent to $(n+1)(6n-1)^2 \geq 36n^3$, or $24n^2 - 11n + 1 \geq 0$. But this last inequality is true since $24n^2 - 11n + 1 = (3n-1)(8n-1)$, and the induction is complete.