已知数列 $\left\{a_n\right\}$ 满足：$a_1=\dfrac{\pi}{2} $，$a_{n+1}=a_n-\dfrac{\sin a_n}{n+1}$（$n \in\mathbb N^{\ast}$）．
1、证明：$0<a_{n+1}<a_n \leqslant \dfrac{\pi}{2}$．
2、证明：$n a_n<10$．
解析
1、先证明 $a_n\in\left(0,\dfrac{\pi}2\right]$，考虑递推，有\[0<\dfrac 12a_n=a_n-\dfrac 12a_n<a_{n+1}=a_n-\dfrac{\sin a_n}{n+1}<a_n\leqslant \dfrac{\pi}2,\]有界性得证．进而易得单调性，命题得证．
2、根据题意，有\[0<(n+1)a_{n+1}-na_n=a_n-\sin a_n<\dfrac 16a_n^3,\]因此 $\{na_n\}$ 单调递增，且 $\{a_n\}$ 单调递减，考虑证明 $a_n\leqslant \dfrac{p}{\sqrt n}$，其中 $p$ 为正常数．由\[a_{n+1}=a_n-\dfrac{\sin a_n}{n+1}<\dfrac{na_n+\dfrac 16a_n^3}{n+1},\]为了保证递推成立，只需要\[\dfrac{n\cdot \dfrac {p}{\sqrt n}+\dfrac 16\left( \dfrac {p}{\sqrt n}\right)^3}{n+1}\leqslant \dfrac {p}{\sqrt {n+1}},\]即\[\sqrt n+\dfrac {p^2}{6n\sqrt n}\leqslant \sqrt{n+1}\iff p^2<\dfrac{6n\sqrt n}{\sqrt{n+1}+\sqrt n},\]而右侧关于 $n$ 单调递增． 当起点选为 $n=1$ 时，有\[\dfrac{\pi}2=a_1\leqslant p\leqslant \sqrt{6\left(\sqrt 2-1\right)},\]因此取 $p=a_1=\dfrac{\pi-1}2$ 即可． 这样就有\[\begin{split} (n+1)a_{n+1}&<a_1+\dfrac 16\sum_{k=1}^n\left(\dfrac{a_1}{\sqrt k}\right)^3\\ &<a_1+\dfrac16a_1^3\sum_{k=1}^n\left(\dfrac{2}{\sqrt {k-\frac 12}}-\dfrac{2}{\sqrt{k+\frac 12}}\right)\\ &=a_1+\dfrac 13a_1^3\left(2\sqrt 2-\dfrac{2}{\sqrt{n+\frac 12}}\right)\\ &<a_1+\dfrac {2\sqrt 2}3a_1^3\\ &<10,\end{split}\]命题得证．
备注 事实上，有 $a_1+\dfrac {2\sqrt 2}3a_1^3=5.2249\cdots$．也可以利用\[\dfrac{1}{na_n}-\dfrac{1}{(n+1)a_{n+1}}=\dfrac{(n+1)a_{n+1}-na_n}{n(n+1)a_na_{n+1}}<\dfrac{a_n^2}{6n(n+1)a_{n+1}}<\dfrac{a_n^2}{6n\cdot na_n}=\dfrac{a_n}{6n^2}\leqslant \dfrac{\pi}{12n^2},\]于是\[\sum_{k=1}^n\left(\dfrac{1}{ka_k}-\dfrac{1}{(k+1)a_{k+1}}\right)\leqslant \dfrac{\pi}{12}\sum_{k=1}^n\dfrac{1}{k^2}<\dfrac{\pi}{12}\left(1+\sum_{k=2}^n\left(\dfrac{1}{k-\frac 12}-\dfrac{1}{k+\frac 12}\right)\right)=\dfrac{5\pi}{36},\]因此\[\dfrac{1}{(n+1)a_{n+1}}> \dfrac{1}{a_1}-\dfrac{5\pi}{36}\implies (n+1)a_{n+1} <\dfrac{36\pi}{72-5\pi^2}<10,\]其中 $\dfrac{36\pi}{72-5\pi^2}=4.9928\cdots$．