求证：$\displaystyle\prod\limits_{k=1}^n {\left( {1+\dfrac{1}{{{9^k}}}} \right)<{\rm{2}}} $．
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【答案】
略
【解析】
根据题意，有\[\begin{split}
\dfrac{1}{{\prod\limits_{k=1}^n {\left( {1+\dfrac{1}{{{9^k}}}} \right)} }} &= \prod\limits_{k=1}^n {\dfrac{1}{{1+\dfrac{1}{{{9^k}}}}}}=\prod\limits_{k=1}^n {\left( {1-\dfrac{1}{{{9^k}+1}}} \right)} \\
&> 1-\sum\limits_{k=1}^n {\dfrac{1}{{{9^k}+1}}}>1-\sum\limits_{k=1}^n {\dfrac{1}{{{9^k}}}} \\
&> 1-\dfrac{{\dfrac{1}{9}}}{{1-\dfrac{1}{9}}}=\dfrac{7}{8},\end{split}\]于是$$\displaystyle\prod\limits_{k=1}^n {\left( {1+\dfrac{1}{{{9^k}}}} \right)}<\dfrac{8}{7}<2.$$
\dfrac{1}{{\prod\limits_{k=1}^n {\left( {1+\dfrac{1}{{{9^k}}}} \right)} }} &= \prod\limits_{k=1}^n {\dfrac{1}{{1+\dfrac{1}{{{9^k}}}}}}=\prod\limits_{k=1}^n {\left( {1-\dfrac{1}{{{9^k}+1}}} \right)} \\
&> 1-\sum\limits_{k=1}^n {\dfrac{1}{{{9^k}+1}}}>1-\sum\limits_{k=1}^n {\dfrac{1}{{{9^k}}}} \\
&> 1-\dfrac{{\dfrac{1}{9}}}{{1-\dfrac{1}{9}}}=\dfrac{7}{8},\end{split}\]于是$$\displaystyle\prod\limits_{k=1}^n {\left( {1+\dfrac{1}{{{9^k}}}} \right)}<\dfrac{8}{7}<2.$$
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