已知 ${a_1}=1$，${a_{n+1}}=\left( {1+\dfrac{1}{{{2^n}}}} \right){a_n}+\dfrac{1}{{{n^2}}}$，求证：${a_n}<{{\rm{e}}^{\frac{{11}}{4}}}$．
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【答案】
略
【解析】
不难证明 ${a_n}$ 单调递增，于是 ${a_n} \geqslant 1$，进而$$\ln \dfrac{{{a_{n+1}}}}{{{a_n}}}=\ln \left( {1+\dfrac{1}{{{2^n}}}+\dfrac{1}{{{n^2}}} \cdot \dfrac{1}{{{a_n}}}} \right) \leqslant \ln \left( {1+\dfrac{1}{{{2^n}}}+\dfrac{1}{{{n^2}}}} \right)
< \dfrac{1}{{{2^n}}}+\dfrac{1}{{{n^2}}},$$于是\[\begin{split}\ln {a_{n+1}}&= \sum\limits_{k=1}^n {\ln \dfrac{{{a_{k+1}}}}{{{a_k}}}}
< \sum\limits_{k=1}^n {\left( {\dfrac{1}{{{2^n}}}+\dfrac{1}{{{n^2}}}} \right)} \\
&< \dfrac{{\dfrac{1}{2}}}{{1-\dfrac{1}{2}}}+\dfrac{1}{{{1^2}}}+\dfrac{1}{{{2^2}}}+\sum\limits_{k=3}^n {\left( {\dfrac{1}{{n-1}}-\dfrac{1}{n}} \right)} \\
&< 1+1+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{{11}}{4}.\end{split}\]
< \dfrac{1}{{{2^n}}}+\dfrac{1}{{{n^2}}},$$于是\[\begin{split}\ln {a_{n+1}}&= \sum\limits_{k=1}^n {\ln \dfrac{{{a_{k+1}}}}{{{a_k}}}}
< \sum\limits_{k=1}^n {\left( {\dfrac{1}{{{2^n}}}+\dfrac{1}{{{n^2}}}} \right)} \\
&< \dfrac{{\dfrac{1}{2}}}{{1-\dfrac{1}{2}}}+\dfrac{1}{{{1^2}}}+\dfrac{1}{{{2^2}}}+\sum\limits_{k=3}^n {\left( {\dfrac{1}{{n-1}}-\dfrac{1}{n}} \right)} \\
&< 1+1+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{{11}}{4}.\end{split}\]
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