证明：$\displaystyle \sum\limits_{k=1}^n {\dfrac{1}{{{k^4}}}}<\dfrac{{53}}{{45}}$．
【难度】
【出处】
无
【标注】
【答案】
略
【解析】
利用\[\begin{split}\dfrac{1}{{{n^4}}} &< \dfrac{1}{{\left( {n-\dfrac{3}{2}} \right)\left( {n-\dfrac{1}{2}} \right)\left( {n+\dfrac{1}{2}} \right)\left( {n+\dfrac{3}{2}} \right)}}\\
&= \dfrac{1}{3}\left[ {\dfrac{1}{{\left( {n-\dfrac{3}{2}} \right)\left( {n-\dfrac{1}{2}} \right)\left( {n+\dfrac{1}{2}} \right)}}-\dfrac{1}{{\left( {n-\dfrac{1}{2}} \right)\left( {n+\dfrac{1}{2}} \right)\left( {n+\dfrac{3}{2}} \right)}}} \right].\end{split}\]
&= \dfrac{1}{3}\left[ {\dfrac{1}{{\left( {n-\dfrac{3}{2}} \right)\left( {n-\dfrac{1}{2}} \right)\left( {n+\dfrac{1}{2}} \right)}}-\dfrac{1}{{\left( {n-\dfrac{1}{2}} \right)\left( {n+\dfrac{1}{2}} \right)\left( {n+\dfrac{3}{2}} \right)}}} \right].\end{split}\]
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