Let $a_1=2$ and $a_i=2\sin\frac{a_{i-1}}2$ for $i\ge2$. Prove that $\sum\limits_{i=1}^{2024}a_i<314$.
In fact, $\sum\limits_{i=1}^{2024}a_i\approx298.796$, so the inequality is very strong.
I tried to establish inequalities with Taylor's series, and got $\sum\limits_{i=1}^{2024}a_i<582$. This is not enough. We need better (more accurate) ways to estimate the series.
This question is from the "$\pi$ day math contest" of THU, which has ended.
Best Answer
The following is strongly inspired by JimmyK4542's comment and Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$.
Set $b_n = a_n/2$, then $b_1 = 1$ and $b_{n+1} = \sin(b_n)$. In Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ it is shown that $$ b_n \sim \sqrt{\frac{3}{n}} \text{ for } n \to \infty \, . $$ This suggests that an explicit extimate of the form $$ \tag{1} b_n \le \sqrt{\frac{3}{n+2}} $$ might hold for all $n \ge 1$. If that is true then $$ \begin{align} \sum_{n=1}^{2024} a_n &\le 2 \sum_{n=1}^{2024} \sqrt{\frac{3}{n+2}} = 2 \sqrt 3 \sum_{n=3}^{2026} \frac{1}{\sqrt n} \\ &< 2 \sqrt 3 \int_2^{2026} \frac{dx}{\sqrt x} = 4 \sqrt 3 (\sqrt{2026} - \sqrt 2) \\ &\approx 302.048 < 314 \, . \end{align} $$
In order to prove $(1)$ by induction we need to show that $$ \sin \sqrt{\frac 3k} \le \sqrt{\frac{3}{k+1}} \text{ for } k \ge 3 \, . $$ With the substitution $x = \sqrt{3/k}$ this is equivalent to $$ \tag{2} \sin^2(x) \le \frac{3x^3}{3 + x^2} \text{ for } 0 < x \le 1 $$ or $$ \tag{3} \frac{1}{\sin^2(x)} > \frac{1}{x^2} + \frac 13 \text{ for } 0 < x \le 1 \, . $$
I will now prove inequality $(3)$, but I wonder if there is a simpler way.
For $0 < x < \pi$ is (see e.g. here) $$ \frac{1}{\sin(x)} = \csc(x) =\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}2\left(2^{2n-1}-1\right)B_{2n}}{(2n)!}}x^{2n-1} \\ = \frac 1x + \frac{1}{6} x +\frac{7}{360}x^{3}+\frac{31}{15120}x^{5}+\cdots $$ where $B_n$ are the Bernoulli numbers. $B_{2n}$ is negative if $n$ is even, and positive otherwise, so that $(-1)^{n+1}B_{2n}$ is positive, i.e. all non-zero coefficients in that series are positive.
It follows that for $0 < x < \pi$ $$ \frac{1}{\sin^2(x)} > \left(\frac 1x + \frac x6\right)^2 > \frac{1}{x^2} + \frac 13 $$ and that completes the proof.