设数列 $\left\{ {{a_n}} \right\}$ 满足 ${a_1} = 0$ 且 $\dfrac{1}{{1 - a{}_{n + 1}}} - \dfrac{1}{{1 - a{}_n}} = 1.$
【难度】
【出处】
2011年高考大纲全国卷（理）
【标注】
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求 $\left\{ {a{}_n} \right\}$ 的通项公式；标注答案解析由题设\[\dfrac{1}{{1 - a{}_{n + 1}}} - \dfrac{1}{{1 - a{}_n}} = 1 , \dfrac{1}{{1 - a{}_1}}{\text{ = }}1,\]即 $\left\{ {\dfrac{1}{{1 - a{}_n}}} \right\}$ 是首项为 $1$，公差为 $ 1 $ 的等差数列．故\[\dfrac{1}{{1 - a{}_n}}{\text{ = }}n.\]所以\[{a_n} = 1 - \dfrac{1}{n}.\]
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设 ${b_n} = \dfrac{{1 - \sqrt {{a_{n + 1}}} }}{{\sqrt n }}$，记 $\displaystyle {S_n} = \sum\limits_{k = 1}^n {{b_k}} $，证明：${S_n} < 1$．标注答案解析由（1）得\[\begin{split}{b_n} & = \dfrac{{1 - \sqrt {{a_{n + 1}}} }}{{\sqrt n }} \\& = \dfrac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt {n + 1}\cdot \sqrt n }} \\& = \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }},\end{split}\]所以\[\begin{split}{S_n} & = \sum\limits_{k = 1}^n {{b_k}} \\& = \sum\limits_{k = 1}^n {\left( {\dfrac{1}{{\sqrt k }} - \dfrac{1}{{\sqrt {k + 1} }}} \right) } \\& = 1 - \dfrac{1}{{\sqrt {n + 1} }} \\& < 1.\end{split}\]
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