数列 $\left\{ {{a_n}} \right\}$ 中 ${a_1}=2$，${a_{n+1}}=a_n^2-{a_n}+1$．
求证：$$1-\dfrac{1}{{{{2014}^{2014}}}}<\sum\limits_{k=1}^{2014} {\dfrac{1}{{{a_k}}}}<1.$$
求证：$$1-\dfrac{1}{{{{2014}^{2014}}}}<\sum\limits_{k=1}^{2014} {\dfrac{1}{{{a_k}}}}<1.$$
【难度】
【出处】
无
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【答案】
略
【解析】
根据题意\[\begin{split}
&\quad {a_{n+1}}-1={a_n}\left( {{a_n}-1} \right) \\
& \Rightarrow \dfrac{1}{{{a_{n+1}}-1}}=\dfrac{1}{{{a_n}-1}}-\dfrac{1}{{{a_n}}} \\
& \Rightarrow \sum\limits_{k=1}^n {\dfrac{1}{{{a_k}}}}=\sum\limits_{k=1}^n {\left( {\dfrac{1}{{{a_k}-1}}-\dfrac{1}{{{a_{k+1}}-1}}} \right)}=\dfrac{1}{{{a_1}-1}}-\dfrac{1}{{{a_{n+1}}-1}}=1-\dfrac{1}{{{a_{n+1}}-1}}
\end{split}\]于是原不等式即$$ {a_{2015}}-1>{2014^{2014}}$$考虑用数学归纳法，若 ${a_{n+1}}-1>{n^n}$，则$${a_{n+2}}-1={a_{n+1}}\left( {{a_{n+1}}-1} \right)>{n^n}\left( {{n^n}-1} \right)$$于是只需要证明$${n^n}\left( {{n^n}-1} \right)>{\left( {n+1} \right)^{n+1}},$$即$$\dfrac{{{n^n}-1}}{{n+1}}>{\left( {1+\dfrac{1}{n}} \right)^n}.$$将 $\dfrac{{{n^n}-1}}{{n+1}}$ 放缩至$$\dfrac{{{n^3}-1}}{{n+1}}={n^2}-n+1(n \geqslant 3)\geqslant 7>{\rm{3}}>{\left( {1+\dfrac{1}{n}} \right)^n}$$因此选择归纳起点为 $n=3$ 时即可．
事实上，${a_2}=3$，${a_3}=7$，于是 ${a_3}-1>{2^2}$ 成立．
&\quad {a_{n+1}}-1={a_n}\left( {{a_n}-1} \right) \\
& \Rightarrow \dfrac{1}{{{a_{n+1}}-1}}=\dfrac{1}{{{a_n}-1}}-\dfrac{1}{{{a_n}}} \\
& \Rightarrow \sum\limits_{k=1}^n {\dfrac{1}{{{a_k}}}}=\sum\limits_{k=1}^n {\left( {\dfrac{1}{{{a_k}-1}}-\dfrac{1}{{{a_{k+1}}-1}}} \right)}=\dfrac{1}{{{a_1}-1}}-\dfrac{1}{{{a_{n+1}}-1}}=1-\dfrac{1}{{{a_{n+1}}-1}}
\end{split}\]于是原不等式即$$ {a_{2015}}-1>{2014^{2014}}$$考虑用数学归纳法，若 ${a_{n+1}}-1>{n^n}$，则$${a_{n+2}}-1={a_{n+1}}\left( {{a_{n+1}}-1} \right)>{n^n}\left( {{n^n}-1} \right)$$于是只需要证明$${n^n}\left( {{n^n}-1} \right)>{\left( {n+1} \right)^{n+1}},$$即$$\dfrac{{{n^n}-1}}{{n+1}}>{\left( {1+\dfrac{1}{n}} \right)^n}.$$将 $\dfrac{{{n^n}-1}}{{n+1}}$ 放缩至$$\dfrac{{{n^3}-1}}{{n+1}}={n^2}-n+1(n \geqslant 3)\geqslant 7>{\rm{3}}>{\left( {1+\dfrac{1}{n}} \right)^n}$$因此选择归纳起点为 $n=3$ 时即可．
事实上，${a_2}=3$，${a_3}=7$，于是 ${a_3}-1>{2^2}$ 成立．
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