求证：$\displaystyle\dfrac{{\ln {2^2}}}{{{2^2}}}+\dfrac{{\ln {3^2}}}{{{3^2}}}+\dfrac{{\ln {4^2}}}{{{4^2}}}+\cdots+\dfrac{{\ln {n^2}}}{{{n^2}}}>\dfrac{{n-1}}{{2n\left( {n+1} \right)}}$（$n \geqslant 2$）．
【难度】
【出处】
无
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【答案】
略
【解析】
放缩裂项，有\[\begin{split}
\sum\limits_{k=2}^n {\dfrac{{\ln {k^2}}}{{{k^2}}}} &>\ln 4 \cdot \sum\limits_{k=2}^n {\dfrac{1}{{{k^2}}}}
\\
&>\ln 4 \cdot \sum\limits_{k=2}^n {\left( {\dfrac{1}{k}-\dfrac{1}{{k+1}}} \right)} \\
&=\ln 4\left( {\dfrac{1}{2}-\dfrac{1}{{n+1}}} \right)\\
&= \dfrac{{n-1}}{{2\left( {n+1} \right)}} \cdot \ln 4\\
&> \dfrac{{n-1}}{{2\left( {n+1} \right)}} \cdot \dfrac{1}{n}.\end{split}\]
\sum\limits_{k=2}^n {\dfrac{{\ln {k^2}}}{{{k^2}}}} &>\ln 4 \cdot \sum\limits_{k=2}^n {\dfrac{1}{{{k^2}}}}
\\
&>\ln 4 \cdot \sum\limits_{k=2}^n {\left( {\dfrac{1}{k}-\dfrac{1}{{k+1}}} \right)} \\
&=\ln 4\left( {\dfrac{1}{2}-\dfrac{1}{{n+1}}} \right)\\
&= \dfrac{{n-1}}{{2\left( {n+1} \right)}} \cdot \ln 4\\
&> \dfrac{{n-1}}{{2\left( {n+1} \right)}} \cdot \dfrac{1}{n}.\end{split}\]
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