设 $a_k=\dfrac 1{k^2}+\dfrac 1{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac 1{(k+1)^2-1}$，求证：$2011\in \left(\dfrac 2{a_{2010}},\dfrac 2{a_{2011}}\right)$．
【难度】
【出处】
2011年湖南省高中数学竞赛
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【答案】
略
【解析】
易知 $a_k$ 的表达式共有 $2k+1$ 项，分别考察其前 $k$ 项的和与后 $k+1$ 项的和．
因为$$\dfrac 1{k^2}+\dfrac 1{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac 1{k^2+k-1}>\dfrac k{k^2+k}=\dfrac 1{k+1},$$又$$\dfrac 1{k^2}+\dfrac{1}{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac 1{k^2+k-1}<\dfrac k{k^2}=\dfrac 1k,$$所以$$\dfrac 1{k+1}<\dfrac 1{k^2}+\dfrac 1{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac{1}{k^2+k-1}<\dfrac 1k\cdots {\text{ ① }}$$同理可证$$\dfrac 1{k+1}<\dfrac 1{k^2+k}+\dfrac 1{k^2+k+1}+\dfrac 1{k^2+k+2}+\cdots +\dfrac 1{(k+1)^2-1}<\dfrac {1}{k}\cdots {\text{ ② }}$$由 ① $+$ ②，可得$$\dfrac 2{k+1}<a_k<\dfrac 2k,$$由此得$$\dfrac 1{a_k}<\dfrac{k+1}{2}<\dfrac 1{a_{k+1}}.$$取 $k=2010$，得$$\dfrac 1{a_{2010}}<\dfrac{2011}{2}<\dfrac 1{a_{2011}},$$即$$\dfrac 2{a_{2010}}<2011<\dfrac 2{a_{2011}},$$所以 $2011\in\left(\dfrac 2{a_{2010}},\dfrac 2{a_{2011}}\right)$．
因为$$\dfrac 1{k^2}+\dfrac 1{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac 1{k^2+k-1}>\dfrac k{k^2+k}=\dfrac 1{k+1},$$又$$\dfrac 1{k^2}+\dfrac{1}{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac 1{k^2+k-1}<\dfrac k{k^2}=\dfrac 1k,$$所以$$\dfrac 1{k+1}<\dfrac 1{k^2}+\dfrac 1{k^2+1}+\dfrac 1{k^2+2}+\cdots +\dfrac{1}{k^2+k-1}<\dfrac 1k\cdots {\text{ ① }}$$同理可证$$\dfrac 1{k+1}<\dfrac 1{k^2+k}+\dfrac 1{k^2+k+1}+\dfrac 1{k^2+k+2}+\cdots +\dfrac 1{(k+1)^2-1}<\dfrac {1}{k}\cdots {\text{ ② }}$$由 ① $+$ ②，可得$$\dfrac 2{k+1}<a_k<\dfrac 2k,$$由此得$$\dfrac 1{a_k}<\dfrac{k+1}{2}<\dfrac 1{a_{k+1}}.$$取 $k=2010$，得$$\dfrac 1{a_{2010}}<\dfrac{2011}{2}<\dfrac 1{a_{2011}},$$即$$\dfrac 2{a_{2010}}<2011<\dfrac 2{a_{2011}},$$所以 $2011\in\left(\dfrac 2{a_{2010}},\dfrac 2{a_{2011}}\right)$．
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