求证：$\displaystyle\prod\limits_{n=1}^\infty {\left( {1-\dfrac{1}{{{4^n}}}} \right)}>\dfrac{2}{3}$．
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【答案】
略
【解析】
利用等比放缩，有\[\begin{split}
\ln \prod\limits_{n=1}^\infty {\left( {1-\dfrac{1}{{{4^n}}}} \right)} &= \sum\limits_{n=1}^\infty {\ln \left( {1-\dfrac{1}{{{4^n}}}} \right)}=\ln \dfrac{3}{4}+\sum\limits_{n=2}^\infty {\ln \left( {1-\dfrac{1}{{{4^n}}}} \right)} \\
&>\ln \dfrac{3}{4}+\sum\limits_{n=2}^\infty {\dfrac{{-\dfrac{1}{{{4^n}}}}}{{1-\dfrac{1}{{{4^n}}}}}}=\ln \dfrac{3}{4}-\sum\limits_{n=2}^\infty {\dfrac{1}{{{4^n}-1}}} \\
&>\ln \dfrac{3}{4}-\dfrac{{\dfrac{1}{{{4^2}-1}}}}{{1-\dfrac{1}{4}}}=\ln \dfrac{3}{4}-\dfrac{1}{{20}}>\ln \dfrac{2}{3}.\end{split}\]
\ln \prod\limits_{n=1}^\infty {\left( {1-\dfrac{1}{{{4^n}}}} \right)} &= \sum\limits_{n=1}^\infty {\ln \left( {1-\dfrac{1}{{{4^n}}}} \right)}=\ln \dfrac{3}{4}+\sum\limits_{n=2}^\infty {\ln \left( {1-\dfrac{1}{{{4^n}}}} \right)} \\
&>\ln \dfrac{3}{4}+\sum\limits_{n=2}^\infty {\dfrac{{-\dfrac{1}{{{4^n}}}}}{{1-\dfrac{1}{{{4^n}}}}}}=\ln \dfrac{3}{4}-\sum\limits_{n=2}^\infty {\dfrac{1}{{{4^n}-1}}} \\
&>\ln \dfrac{3}{4}-\dfrac{{\dfrac{1}{{{4^2}-1}}}}{{1-\dfrac{1}{4}}}=\ln \dfrac{3}{4}-\dfrac{1}{{20}}>\ln \dfrac{2}{3}.\end{split}\]
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