求证：$\displaystyle\sum\limits_{k=1}^n {\left( {\dfrac{{{3^k}}}{{{3^k}+1}}+\dfrac{{{3^k}}}{{{3^k}-\dfrac{1}{3}}}} \right)}>2n-\dfrac{1}{4}$．
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【答案】
略
【解析】
原不等式即$$ \sum\limits_{k=1}^n {\left( {\dfrac{1}{{{3^k}+1}}-\dfrac{{\dfrac{1}{3}}}{{{3^k}-\dfrac{1}{3}}}} \right)}<\dfrac{1}{4},$$也即$$ \sum\limits_{k=1}^n {\left( {\dfrac{1}{{{3^k}+1}}-\dfrac{1}{{{3^{k+1}}-1}}} \right)}<\dfrac{1}{4},$$而\[\begin{split}\sum\limits_{k=1}^n {\left( {\dfrac{1}{{{3^k}+1}}-\dfrac{1}{{{3^{k+1}}-1}}} \right)} &< \dfrac{1}{{3+1}}-\dfrac{1}{{{3^2}-1}}+\sum\limits_{k=2}^n {\left( {\dfrac{1}{{{3^k}-1}}-\dfrac{1}{{{3^{k+1}}-1}}} \right)} \\
&= \dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{{{3^{n+1}}-1}}\\
&<\dfrac{1}{4},\end{split}\]因此原不等式成立．
&= \dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{{{3^{n+1}}-1}}\\
&<\dfrac{1}{4},\end{split}\]因此原不等式成立．
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