已知数列 $\{a_{n}\}$ 中，$a_{1}=1$，$a_{2}=\dfrac{1}{4}$，且 $a_{n+1}=\dfrac{(n-1)a_{n}}{n-a_{n}}$，$n=2,3,4,\cdots$，
【难度】
【出处】
2018年全国高中数学联赛辽宁省预赛
【标注】
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求数列 $\{a_{n}\}$ 的通项公式标注答案$a_{n}=\dfrac{1}{3n-2}$解析由已知，对 $n\geqslant 2$ 有，$\dfrac{1}{a_{n+1}}=\dfrac{n-a_{n}}{(n-1)a_{n}}=\dfrac{n}{(n-1)a_{n}}-\dfrac{1}{n-1}$，两边同时除以 $n$，得 $\dfrac{1}{na_{n+1}}=\dfrac{1}{(n-1)a_{n}}-\dfrac{1}{n(n-1)}$，即 $\dfrac{1}{n{{a}_{n}}}-\dfrac{1}{\left(n-1 \right){{a}_{n}}}\text{=}-\left( \dfrac{1}{n-1}-\dfrac{1}{n} \right)$，于是 $\displaystyle \sum\limits_{k\text{=}2}^{n-1}{\left[\dfrac{1}{k{{a}_{k+1}}}-\dfrac{1}{\left( k-1 \right){{a}_{k}}}\right]}\text{=}-\sum\limits_{k\text{=}2}^{n-1}{\left(\dfrac{1}{k-1}-\dfrac{1}{k} \right)}\text{=}-\left( 1-\dfrac{1}{n-1} \right)$，即 $\dfrac{1}{\left(n-1 \right){{a}_{n}}}-\dfrac{1}{{{a}_{2}}}\text{=}-\left( 1-\dfrac{1}{n-1}\right)\text{,}n\geqslant 2$，所以 $\dfrac{1}{\left(n-1 \right){{a}_{n}}}\text{=}\dfrac{1}{{{a}_{2}}}-\left( 1-\dfrac{1}{n-1}\right)\text{=}\dfrac{3n-2}{n-1}\text{,}{{a}_{n}}\text{=}\dfrac{1}{3n-2}\text{,}n\geqslant 2$．又 $n=1$ 时也成立，故 $a_{n}=\dfrac{1}{3n-2}$，$n\in \mathbf N^{*}$．
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求证：对一切 $n\in\mathbf N^{*}$，有 $\displaystyle \sum_{k=1}^{n}a^2_{k}<\dfrac{7}{6}$标注答案略解析当 $k\geqslant 2$ 时，有 $a_{k}^{2}\text{=}\dfrac{1}{{{\left(3k-2 \right)}^{2}}}<\dfrac{1}{\left( 3k-4 \right)\left( 3k-1\right)}\text{=}\dfrac{1}{3}\left( \dfrac{1}{3k-4}-\dfrac{1}{3k-1} \right)$，所以 $n\geqslant 2$ 时，有 $\displaystyle \sum\limits_{k\text{=}1}^{n}{a_{k}^{2}}\text{=}1+\sum\limits_{k\text{=}2}^{n}{a_{k}^{2}}<1+\dfrac{1}{3}\left[\left( \dfrac{1}{2}-\dfrac{1}{5} \right)+\left( \dfrac{1}{5}-\dfrac{1}{8}\right)+\cdots +\left( \dfrac{1}{3n-4}-\dfrac{1}{3n-1} \right)\right]\text{=}1+\dfrac{1}{3}\left( \dfrac{1}{2}-\dfrac{1}{3n-1}\right)<1+\dfrac{1}{6}\text{=}\dfrac{7}{6}$．又 $n=1$ 时，$a^2_{1}=1<\dfrac{7}{6}$．故对一切 $n\in\mathbf N^{\ast}$，有 $\displaystyle\sum^n_{k=1}a^2_k<\dfrac{7}{6}$．
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