1673

已知数列 ${\displaystyle \{a_n\}}$ 和 ${\displaystyle \{{\theta }_n\}}$, ${\displaystyle S_n}$ 为数列 ${\displaystyle \{a_n\}}$ 的前 ${\displaystyle n}$ 项和, ${\displaystyle a_1=1}$, ${\displaystyle {\theta }_n\in (0,\frac{\pi }{2})}$, 且 ${\displaystyle S_n=\frac{1}{2}+\frac{\sqrt{3}}{2}\tan {\theta }_n}$, ${\displaystyle a_{n+1}=\frac{\sqrt{3}}{2\cos {\theta }_n}}$.

1. 求 ${\displaystyle a_2,a_3}$;
2. 求数列 ${\displaystyle \{a_n\}}$ 的通项公式;
3. 求证: ${\displaystyle S_n\le 2+\sqrt{3}(2^{n-2}-1)}$.

1. 解 由 ${\displaystyle S_1=\frac{1}{2}+\frac{\sqrt{3}}{2}\tan {\theta }_1=a_1=1}$, 得 ${\displaystyle {\theta }_1=\frac{\pi }{6}}$, 所以 ${\displaystyle a_2=\frac{\sqrt{3}}{2\cos {\theta }_1}=1}$.
   再由 ${\displaystyle S_2=\frac{1}{2}+\frac{\sqrt{3}}{2}\tan {\theta }_2=a_1+a_2=2}$, 得 ${\displaystyle {\theta }_2=\frac{\pi }{3}}$, 所以 ${\displaystyle a_3=\frac{\sqrt{3}}{2\cos {\theta }_2}=\sqrt{3}}$.
2. 解 因为 $$a_{n+1}=S_{n+1}-S_n=\frac{\sqrt{3}}{2}(\tan {\theta }_{n+1}-\tan {\theta }_n)=\frac{\sqrt{3}}{2\cos {\theta }_n},$$ 所以 $$\tan {\theta }_{n+1}=\tan {\theta }_n+\frac{1}{\cos {\theta }_n}=\frac{2\tan \frac{{\theta }_n}{2}}{1-{\tan }^2\frac{{\theta }_n}{2}}+\frac{1+{\tan }^2\frac{{\theta }_n}{2}}{1-{\tan }^2\frac{{\theta }_n}{2}}=\frac{1+\tan \frac{{\theta }_n}{2}}{1-\tan \frac{{\theta }_n}{2}}=\tan (\frac{{\theta }_n}{2}+\frac{\pi }{4}).$$ 而 ${\displaystyle {\theta }_{n+1},\frac{{\theta }_n}{2}+\frac{\pi }{4}\in (0,\frac{\pi }{2})}$, 所以 $${\theta }_{n+1}=\frac{1}{2}{\theta }_n+\frac{\pi }{4}\Rightarrow {\theta }_{n+1}-\frac{\pi }{2}=\frac{1}{2}({\theta }_n-\frac{\pi }{2}).$$ 这样 ${\displaystyle \{{\theta }_n-\frac{\pi }{2}\}}$ 是以 ${\displaystyle -\frac{\pi }{3}}$ 为首项, ${\displaystyle \frac{1}{2}}$ 为公比的等比数列, 所以 ${\displaystyle {\theta }_n=\frac{\pi }{2}-\frac{\pi }{3}{\left(\frac{1}{2}\right)}^{n-1}}$, 进而 $$a_n=\frac{\sqrt{3}}{2\cos {\theta }_{n-1}}=\frac{\sqrt{3}}{2\sin \left(\frac{\pi }{3\cdot 2^{n-2}}\right)}.$$ (需要单独验证 ${\displaystyle a_1=1}$ 也满足这个表达式).
3. 证明 当 ${\displaystyle n=1}$ 或 ${\displaystyle 2}$ 时, 直接验证不等式成立. 当 ${\displaystyle n\ge 3}$, 只需证明 ${\displaystyle a_n\le \sqrt{3}\cdot 2^{n-3}}$, 这样 $$S_n=2+\sum _{i=3}^n{a}_i\le 2+\sum _{i=0}^{n-3}\sqrt{3}\cdot 2^i=2+\sqrt{3}(2^{n-2}-1).$$ 而这等价于证明 ${\displaystyle \sin \left(\frac{\pi }{3\cdot 2^{n-2}}\right)\ge \frac{1}{2^{n-2}}}$. 为此, 构造 ${\displaystyle f(x)=\sin x-\frac{3x}{\pi }}$, ${\displaystyle x\in (0,\frac{\pi }{6}]}$, 则 ${\displaystyle f^{\prime }(x)=\cos x-\frac{3}{\pi }}$, ${\displaystyle f^{″}(x)=-\sin x<0}$; 而 ${\displaystyle f^{\prime }(0)=1-\frac{3}{\pi }>0}$, ${\displaystyle f^{\prime }\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}-\frac{3}{\pi }<0}$, 所以 ${\displaystyle f^{\prime }(x)}$ 存在唯一零点 ${\displaystyle x_0}$, 则 ${\displaystyle f(x)}$ 在 ${\displaystyle (0,x_0)}$ 上增, 在 ${\displaystyle (x_0,\frac{\pi }{6})}$ 上减, 所以 ${\displaystyle f(x)\ge min\{f(0),f\left(\frac{\pi }{6}\right)\}=0}$, 这样 ${\displaystyle f\left(\frac{\pi }{3\cdot 2^{n-2}}\right)\ge 0}$, 我们就完成了证明.