已知数列 $\{a_{n}\}$ 中，$a_{1}=1,a_{2}=\dfrac{1}{4}$，且 $a_{n+1}=\dfrac{(n-1)a_{n}}{n-a_{n}}$（$n=2,3,4,\cdots$）．
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2010年全国高中数学联赛湖北省预赛
【标注】
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求数列 $\{a_{n}\}$ 的通项公式；标注答案$a_{n}=\dfrac{1}{3n-2},n\in\mathbb N^{*}$解析因为$$a_{1}=1,a_{n+1}=\dfrac{(n-1)a_{n}}{n-a_{n}},n=2,3,4,\cdots,$$所以当 $n\geqslant 2$ 时，\[\dfrac{1}{a_{n+1}}=\dfrac{n-a_{n}}{(n-1)a_{n}}=\dfrac{n}{(n-1)a_{n}}-\dfrac{1}{n-1}.\]两边同时除以 $n$，得\[\dfrac{1}{na_{n+1}}=\dfrac{1}{(n-1)a_{n}}-\dfrac{1}{n(n-1)},\]所以\[b_{n+1}-b_{n}=-\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right),\]故\[\begin{split}b_{n}-b_{2}&=\sum\limits_{k=2}^{n-1}\left[b_{k+1}-b_{k}\right]\\&=-\sum\limits _{k=2}^{n-1}\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)\\&=-\left(1-\dfrac{1}{n-1}\right),\end{split}\]于是\[b_{n}=-\left(1-\dfrac{1}{n-1}\right)+b_{2}=\dfrac{3n-2}{n-1},n\in\mathbb N^{*}.\]因此 $a_{n}=\dfrac{1}{3n-2},n\in\mathbb N^{*}$．
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求证：对一切 $n\in\mathbb N^{*}$，有 $\displaystyle \sum\limits_{k=1}^{n}a_{k}^{2}<\dfrac{7}{6}$．标注答案略解析当 $k\geqslant 2$ 时，\[a_{k}^{2}=\dfrac{1}{(3k-2)^{2}}<\dfrac{1}{(3k-4)(3k-1)}=\dfrac{1}{3}\left(\dfrac{1}{3k-4}-\dfrac{1}{3k-1}\right),\]所以，当 $n\geqslant 2$ 时，\[\begin{split}\sum\limits_{k=1}^{n}a_{k}^{2}&=1+\sum\limits_{k=2}^{n}a_{k}^{2}\\&<1+\dfrac{1}{3}\left[\left(\dfrac{1}{2}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{8}\right)+\cdots+\left(\dfrac{1}{3n-4}-\dfrac{1}{3n-1}\right)\right]\\&=1+\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n-1}\right)\\&<1+\dfrac{1}{6}\\&=\dfrac{7}{6}.\end{split}\]又因为当 $n=1$ 时，$$a_{1}^{2}=1<\dfrac{7}{6},$$所以对一切 $n\in\mathbb N^{*}$，有 $\displaystyle \sum\limits_{k=1}^{n}a_{k}^{2}<\dfrac{7}{6}$．
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