已知数列 $\{a_n\}$ 中，$a_1=1$，$a_2=\dfrac 14$，且 $a_{n+1}=\dfrac {(n-1)a_n}{n-a_n}$（$n=2,3,4,\cdots$）．
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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$$\dfrac {1}{a_{n+1}}=\dfrac {n-a_n}{(n-1)a_n}=\dfrac {n}{(n-1)a_n}-\dfrac 1{n-1},$$两边同除以 $n$ 并整理，得$$\dfrac {1}{na_{n+1}}-\dfrac {1}{(n-1)a_n}=-\left(\dfrac {1}{ n-1 } -\dfrac {1}{n }\right),$$于是$$\sum \limits_{k=2}^{n-1}\left[\dfrac {1}{ka_{k+1}}-\dfrac {1}{(k-1)a_k}\right]=-\sum \limits_{k=2}^{n-1}\left(\dfrac {1}{k-1 } -\dfrac {1}{k}\right)=-\left(1-\dfrac {1}{n-1}\right),$$即$$ \dfrac {1}{(n-1)a_n } -\dfrac {1}{a_2} =-\left(1-\dfrac {1}{n-1}\right),n \geqslant 2.$$所以$$\dfrac {1}{(n-1)a_n } =\dfrac {1}{a_2} -\left(1-\dfrac {1}{n-1}\right)=\dfrac {3n-2}{n-1},$$所以 $a_n=\dfrac {1}{3n-2}$，$n \geqslant 2$．
又当 $n=1$ 时也成立，故 $a_n=\dfrac {1}{3n-2}$，$n \in \mathbb N^*$． -
求证：对一切 $n \in \mathbb N^*$，有 $\displaystyle \sum \limits_{k=1}^{n}a_k^2<\dfrac 76$．标注答案略解析当 $k \geqslant 2$，有\[\begin{split}a_k^2&=\dfrac {1}{(3k-2)^2}\\ &<\dfrac {1}{(3k-4)(3k-1)}\\ &= \dfrac 13 \left(\dfrac {1}{3k-4}-\dfrac {1}{3k-1}\right),\end{split}\]所以，当 $n \geqslant 2$，有\[\begin{split}\sum \limits_{k=1}^{n}a_k^2&=1+\sum \limits_{k=2}^{n}a_k^2 \\& < 1+\dfrac 13 \left[\left(\dfrac 12-\dfrac 15\right)+\left(\dfrac 15-\dfrac 18\right)+\cdots +\left(\dfrac 1{3n-4}-\dfrac 1{3n-1}\right)\right]\\ &= 1+\dfrac 13\left(\dfrac 12-\dfrac 1{3n-1}\right) \\&<1+\dfrac 16=\dfrac 76.\end{split}\]又 $n=1$ 时，$a_1^2=1<\dfrac 76$．故对一切 $n \in \mathbb N^*$，有 $\displaystyle \sum \limits_{k=1}^{n}a_k^2<\dfrac 76$．
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