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[不等式] 2015哈萨克斯坦数学奥林匹克不等式

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longzaifei Posted 2015-1-22 18:07 |Read mode
求证: $ \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{(n+1)^2} < n(1-\frac{1}{\sqrt[n]{2}}) $

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 Author| longzaifei Posted 2015-1-22 18:44
Last edited by longzaifei 2015-1-22 18:58发现$ \dfrac{1}{(n+1)^2} < \dfrac{1}{(n+\frac{1}{4})(n+\frac{5}{4})} $
放缩好像可以

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