设函数 $f\left( x \right)=x-x\ln x$．数列 $\left\{ {{a_n}} \right\}$ 满足：$0<{a_1}<1$，${a_{n+1}}=f\left( {{a_n}} \right)$．设 $b \in \left( {{a_1} , 1} \right)$，整数 $k \geqslant \dfrac{{{a_1}-b}}{{{a_1}\ln b}}$．证明：${a_{k+1}}>b$．
【难度】
【出处】
无
【标注】
【答案】
略
【解析】
由不动点法容易分析出数列 $\left\{ {{a_n}} \right\}$ 单调递增，极限为 $1$．
由于 ${a_1}<b<1$，于是存在正整数 $N$，使得 ${a_N} \leqslant b<{a_{N+1}}$，题意即需要我们去证明 $N \leqslant k$．
由$${a_{n+1}}={a_n}-{a_n}\ln {a_n}$$得$${a_{n+1}}-{a_n}=- {a_n}\ln {a_n}.$$于是\[\begin{split}
{a_{k+1}}&={a_1}+\sum\limits_{m=1}^k {\left( {{a_{m+1}}-{a_m}} \right)}= {a_1}+\sum\limits_{m=1}^k {\left( {-{a_m}\ln {a_m}} \right)} \\
&> {a_1}+\sum\limits_{m=1}^k {\left( {-{a_1}\ln b} \right)} = {a_1}+k\left( {-{a_1}\ln b} \right)\\
& \geqslant {a_1}+\dfrac{{b-{a_1}}}{{-{a_1}\ln b}} \cdot \left( {-{a_1}\ln b} \right)=b.\end{split}\]
由于 ${a_1}<b<1$，于是存在正整数 $N$，使得 ${a_N} \leqslant b<{a_{N+1}}$，题意即需要我们去证明 $N \leqslant k$．
由$${a_{n+1}}={a_n}-{a_n}\ln {a_n}$$得$${a_{n+1}}-{a_n}=- {a_n}\ln {a_n}.$$于是\[\begin{split}
{a_{k+1}}&={a_1}+\sum\limits_{m=1}^k {\left( {{a_{m+1}}-{a_m}} \right)}= {a_1}+\sum\limits_{m=1}^k {\left( {-{a_m}\ln {a_m}} \right)} \\
&> {a_1}+\sum\limits_{m=1}^k {\left( {-{a_1}\ln b} \right)} = {a_1}+k\left( {-{a_1}\ln b} \right)\\
& \geqslant {a_1}+\dfrac{{b-{a_1}}}{{-{a_1}\ln b}} \cdot \left( {-{a_1}\ln b} \right)=b.\end{split}\]
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