设函数 ${f_n}\left( x \right) = - 1 + x + \dfrac{{{x^2}}}{{{2^2}}} + \dfrac{{{x^3}}}{{{3^2}}} + \cdots + \dfrac{{{x^n}}}{{{n^2}}}$（$n=1,2,\cdots$）．
1、证明：对每个 $n \in {{\mathbb{N}}^{\ast}}$，存在唯一的 $x_n\in\left[\dfrac 23,1\right]$，满足 ${f_n}\left( {{x_n}} \right) = 0$．
2、证明：对任意 $p \in {{\mathbb{N}}^{\ast}}$，由 $(1)$ 中 ${x_n}$ 构成的数列 $\left\{ {{x_n}} \right\}$ 满足 $0 < {x_n} - {x_{n + p}} < \dfrac{1}{n}$．
解析
1、本题考查函数的零点，根据零点的存在性定理结合函数的单调性论证即可． 当 $n = 1$ 时，${f_n}\left( {\dfrac{2}{3}} \right) = - \dfrac{1}{3} < 0$； 当 $n \geqslant 2$ 时，根据题意\[\begin{split}{f_n}\left( {\dfrac{2}{3}} \right) &= - 1 + \dfrac{2}{3} + \dfrac{1}{{{2^2}}} \cdot {\left( {\dfrac{2}{3}} \right)^2} + \dfrac{1}{{{3^2}}} \cdot {\left( {\dfrac{2}{3}} \right)^3} + \cdots + \dfrac{1}{{{n^2}}} \cdot {\left( {\dfrac{2}{3}} \right)^n} \\&< - 1 + \dfrac{2}{3} + \dfrac{1}{{{2^2}}}\left[ {{{\left( {\dfrac{2}{3}} \right)}^2} + {{\left( {\dfrac{2}{3}} \right)}^3} + \cdots + {{\left( {\dfrac{2}{3}} \right)}^n}} \right]\\& < - \dfrac{1}{3} + \dfrac{1}{4} \cdot \dfrac{{{{\left( {\dfrac{2}{3}} \right)}^2}}}{{1 - \dfrac{2}{3}}}= 0,\end{split}\] 又 ${f_n}\left( 1 \right) \geqslant 0$，且 ${f_n}\left( x \right)$ 单调递增，因此对每个 $n \in {{\mathbb{N}}^{\ast}}$，存在唯一的 $x_n\in\left[\dfrac 23,1\right]$，满足 ${f_n}\left( {{x_n}} \right) = 0$．
2、本题考查函数的零点与递推数列，利用差分的技巧化简零点的表达式是解决问题的关键． 根据题意 ${f_n}\left( {{x_n}} \right) = 0$，${f_{n + p}}\left( {{x_{n + p}}} \right) = 0$，而$$\forall p \in {\mathbb N^{\ast}} , {f_{n + p}}\left( {{x_n}} \right) = - 1 + \sum\limits_{k = 1}^n {\dfrac{{x_n^k}}{{{k^2}}} + \sum\limits_{k = n + 1}^{n + p} {\dfrac{{x_n^k}}{{{k^2}}}} } = {f_n}\left( {{x_n}} \right) + \sum\limits_{k = n + 1}^{n + p} {\dfrac{{x_n^k}}{{{k^2}}}} > {f_n}\left( {{x_n}} \right) = 0,$$ 于是 $\forall p \in {\mathbb N^{\ast}} , {x_{n + p}} < {x_n}$，即$$\forall p \in {\mathbb N^{\ast}} , {x_n} - {x_{n + p}} > 0.$$ 由 ${f_n}\left( {{x_n}} \right) = 0$ 得$${x_n} = 1 - \sum\limits_{k = 2}^n {\dfrac{{x_n^k}}{{{k^2}}}},$$由 ${f_{n + p}}\left( {{x_{n + p}}} \right) = 0$ 得$${x_{n + p}} = 1 - \sum\limits_{k = 2}^{n + p} {\dfrac{{x_{n + p}^k}}{{{k^2}}}},$$两式相减，有$${x_n} - {x_{n + p}} = \sum\limits_{k = 2}^n {\dfrac{{x_{n + p}^k - x_n^k}}{{{k^2}}}} + \sum\limits_{k = n + 1}^{n + p} {\dfrac{{x_{n + p}^k}}{{{k^2}}}}< \sum\limits_{k = n + 1}^{n + p} {\dfrac{1}{{{k^2}}}} < \dfrac{1}{n} - \dfrac{1}{{n + p}} < \dfrac{1}{n},$$ 综上所述，原命题得证．