947

设数列 ${\displaystyle \{a_n\}}$ 满足 ${\displaystyle a_1=2,a_{n+1}={\displaystyle 2\sin \frac{a_n}{2}}}$，证明: ${\displaystyle {\displaystyle \sum _{n=1}^{2024}{a}_n<314}}$.

记 ${\displaystyle b_n={\displaystyle \frac{a_n}{2}}}$, 则 ${\displaystyle b_1=1,b_{n+1}=\sin b_n}$. 下面用数学归纳法证明 ${\displaystyle b_n<{\displaystyle \sqrt{\frac{3}{n}}}}$. 这对 ${\displaystyle n=1}$ 显然. 事实上只需要证明$$\begin{array}{r}\sin \sqrt{\frac{3}{n}}<\sqrt{\frac{3}{n+1}}⟺2{\sin }^2\sqrt{\frac{3}{n}}=1-\cos 2\sqrt{\frac{3}{n}}<\frac{6}{n+1}.\end{array}$$
做换元 ${\displaystyle t=2\sqrt{\frac{3}{n}}\le 2\sqrt{3}}$, 则要证明 ${\displaystyle g(t)=\cos t+\frac{6t^2}{12+t^2}-1>0}$.
求导得 $$\begin{array}{rl}{g}^{\prime }(t)& =-\sin t+\frac{144t}{(12+t^2{)}^2}\ge -t+\frac{t^3}{6}-\frac{t^5}{120}+\frac{144t}{(12+t^2{)}^2}\\ & =\frac{t^5(216-4t^2-t^4)}{120(12+t^2{)}^2}>0.\end{array}$$ 这里对分子用到了 ${\displaystyle t\le 2\sqrt{3}}$. 从而 ${\displaystyle g(t)>g(0)=0}$.
因此不等式得证! 因此$$\begin{array}{rl}\sum _{n=1}^{2024}{b}_n<& 2\sqrt{3}\sum _{n=1}^{2024}\frac{1}{2\sqrt{n}}<2\sqrt{3}\sum _{n=1}^{2024}\frac{1}{\sqrt{n}+\sqrt{n-1}}=2\sqrt{3}\sum _{n=1}^{2024}(\sqrt{n}-\sqrt{n-1})\\ =& 2\sqrt{3}\cdot \sqrt{2024}<90\sqrt{3}<157,\end{array}$$
从而${\displaystyle {\displaystyle \sum _{n=1}^{2024}{a}_n<314}}$, 得证！