求证：$\displaystyle\sum\limits_{k=1}^n {\left[ {\left( {1+\dfrac{1}{k}} \right)\ln \left( {1+\dfrac{1}{k}} \right)-\dfrac{1}{k}} \right]} <\dfrac 67$．
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【出处】
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【答案】
$\dfrac 67$
【解析】
由于$$\ln \left( {1+\dfrac{1}{n}} \right)<\dfrac{{\dfrac{5}{{2n}}}}{{\dfrac{1}{n}+\dfrac{5}{2}}}=\dfrac{1}{{n+\dfrac{2}{5}}}$$于是\[\begin{split}\left( {1+\dfrac{1}{n}} \right)\ln \left( {1+\dfrac{1}{n}} \right)-\dfrac{1}{n}&=\dfrac{{\dfrac{3}{5}}}{{n\left( {n+\dfrac{2}{5}} \right)}}
< \dfrac{{\dfrac{3}{5}}}{{\left( {n-\dfrac{3}{{10}}} \right)\left( {n+\dfrac{7}{{10}}} \right)}}\\
&= \dfrac{3}{5}\left( {\dfrac{1}{{n-\dfrac{3}{{10}}}}-\dfrac{1}{{n+\dfrac{7}{{10}}}}} \right)\end{split}\]因此$$\sum\limits_{k=1}^n {\left[ {\left( {1+\dfrac{1}{k}} \right)\ln \left( {1+\dfrac{1}{k}} \right)-\dfrac{1}{k}} \right]}<\dfrac{3}{5} \cdot \dfrac{1}{{1-\dfrac{3}{{10}}}}=\dfrac{6}{7}.$$
< \dfrac{{\dfrac{3}{5}}}{{\left( {n-\dfrac{3}{{10}}} \right)\left( {n+\dfrac{7}{{10}}} \right)}}\\
&= \dfrac{3}{5}\left( {\dfrac{1}{{n-\dfrac{3}{{10}}}}-\dfrac{1}{{n+\dfrac{7}{{10}}}}} \right)\end{split}\]因此$$\sum\limits_{k=1}^n {\left[ {\left( {1+\dfrac{1}{k}} \right)\ln \left( {1+\dfrac{1}{k}} \right)-\dfrac{1}{k}} \right]}<\dfrac{3}{5} \cdot \dfrac{1}{{1-\dfrac{3}{{10}}}}=\dfrac{6}{7}.$$
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