实数列 $\{a_n\}$，$a_0=\dfrac12,$ $a_{k+1}=a_k+\dfrac1n a_k^2$（$k\in\mathbb N$），求证：$1-\dfrac1n<a_n<1$．
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【出处】
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【答案】
略
【解析】
对 $1\leqslant k\leqslant n$，我们归纳一个加强的估计$$\dfrac{n+1}{2n-k+2}<a_k<\dfrac{n}{2n-k},$$右边：$$a_{k+1}=\dfrac{a_k}{n}\cdot (n+a_k)<\dfrac1{2n-k}\cdot \dfrac{n(2n-k+1)}{2n-k}<\dfrac{n}{2n-k-1},$$左边：$$\begin{split} a_{k+1}=&a_k+\dfrac1n a_k^2\\
>&\dfrac{n+1}{2n-k+2}+\dfrac1n\left(\dfrac{n+1}{2n-k+2}\right)^2\\
=& \dfrac{n+1}{2n-k+1}+\dfrac{n+1}{(2n-k+2)^2}\cdot\left(\dfrac{n+1}{n}-\dfrac{2n-k+1}{2n-k+2}\right)\\
>&\dfrac{n+1}{2n-k+1}.\end{split}$$
>&\dfrac{n+1}{2n-k+2}+\dfrac1n\left(\dfrac{n+1}{2n-k+2}\right)^2\\
=& \dfrac{n+1}{2n-k+1}+\dfrac{n+1}{(2n-k+2)^2}\cdot\left(\dfrac{n+1}{n}-\dfrac{2n-k+1}{2n-k+2}\right)\\
>&\dfrac{n+1}{2n-k+1}.\end{split}$$
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