设 $\mathbf{N}^{\ast}$ 为正数集合，定义：$a_1=2,a_{n+1}=\min\left\{\lambda|\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_n}+\dfrac{1}{\lambda}<1,\lambda\in\mathbf{N}^{\ast}\right\},n=1,2,\cdots$．求证：$a_{n+1}=a_n^2-a_n+1$．
【难度】
【出处】
2010中国东南数学奥林匹克试题
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【答案】
略
【解析】
由 $a_1=2,a_{2}=\min\left\{\lambda|\dfrac{1}{a_1}+\dfrac{1}{\lambda}<1,\lambda\in\mathbf{N}^{\ast}\right\}$．考虑 $\dfrac{1}{a_1}+\dfrac{1}{\lambda}<1$，则 $\dfrac{1}{\lambda}<1-\dfrac{1}{2}=\dfrac{1}{2},\lambda>2$，从而 $a_2=3$，即 $n=1$ 时，结论成立．
假设对所有 $n\leqslant k-1(k\geqslant 2)$，结论成立．当 $n=k$ 时，由 $a_{k+1}=\min\left\{\lambda|\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}+\dfrac{1}{\lambda}<1,\lambda\in\mathbf{N}^{\ast}\right\}$
考虑 $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}+\dfrac{1}{\lambda}<1$
即 $0<\dfrac{1}{\lambda}<1-\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}\right)$
从而 $\lambda>\dfrac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots-\frac{1}{a_n}}$
下面证明：$\dfrac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots-\frac{1}{a_n}}=a_k(a_k-1)$．
因假设对于 $2\leqslant n\leqslant k,a_n=a_{n-1}(a_{n-1}-1)+1$，有
$\dfrac{1}{a_n-1}=\dfrac{1}{a_{n-1}(a_{n-1}-1)}=\dfrac{1}{a_{n-1}-1}-\dfrac{1}{a_{n-1}}$
所以 $\dfrac{1}{a_{n-1}}=\dfrac{1}{a_{n-1}-1}-\dfrac{1}{a_n-1}$
求和得 $\displaystyle \sum\limits_{i=2}^k\dfrac{1}{a_{i-1}}=1-\dfrac{1}{a_k-1}$，即
$\displaystyle \sum\limits_{i=1}^k\dfrac{1}{a_{i}}=1-\dfrac{1}{a_k-1}+\dfrac{1}{a_k}=1-\dfrac{1}{a_k(a_k-1)}$
于是 $\dfrac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots-\frac{1}{a_k}}=a_k(a_k-1)$
所以 $a_{k+1}=\min\left\{\lambda|\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}+\dfrac{1}{\lambda}<1,\lambda\in\mathbf{N}^{\ast}\right\}=a_k(a_k-1)+1$
由数学归纳法知，对所有正整数 $n$，有 $a_{n+1}=a^2_n-a_n+1$．
假设对所有 $n\leqslant k-1(k\geqslant 2)$，结论成立．当 $n=k$ 时，由 $a_{k+1}=\min\left\{\lambda|\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}+\dfrac{1}{\lambda}<1,\lambda\in\mathbf{N}^{\ast}\right\}$
考虑 $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}+\dfrac{1}{\lambda}<1$
即 $0<\dfrac{1}{\lambda}<1-\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}\right)$
从而 $\lambda>\dfrac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots-\frac{1}{a_n}}$
下面证明：$\dfrac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots-\frac{1}{a_n}}=a_k(a_k-1)$．
因假设对于 $2\leqslant n\leqslant k,a_n=a_{n-1}(a_{n-1}-1)+1$，有
$\dfrac{1}{a_n-1}=\dfrac{1}{a_{n-1}(a_{n-1}-1)}=\dfrac{1}{a_{n-1}-1}-\dfrac{1}{a_{n-1}}$
所以 $\dfrac{1}{a_{n-1}}=\dfrac{1}{a_{n-1}-1}-\dfrac{1}{a_n-1}$
求和得 $\displaystyle \sum\limits_{i=2}^k\dfrac{1}{a_{i-1}}=1-\dfrac{1}{a_k-1}$，即
$\displaystyle \sum\limits_{i=1}^k\dfrac{1}{a_{i}}=1-\dfrac{1}{a_k-1}+\dfrac{1}{a_k}=1-\dfrac{1}{a_k(a_k-1)}$
于是 $\dfrac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots-\frac{1}{a_k}}=a_k(a_k-1)$
所以 $a_{k+1}=\min\left\{\lambda|\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_k}+\dfrac{1}{\lambda}<1,\lambda\in\mathbf{N}^{\ast}\right\}=a_k(a_k-1)+1$
由数学归纳法知，对所有正整数 $n$，有 $a_{n+1}=a^2_n-a_n+1$．
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