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悠闲数学娱乐论坛(第3版)
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2015哈萨克斯坦数学奥林匹克不等式
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2015哈萨克斯坦数学奥林匹克不等式
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longzaifei
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longzaifei
Post time 2015-1-22 18:07
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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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longzaifei
Post time 2015-1-22 18:44
本帖最后由 longzaifei 于 2015-1-22 18:58 编辑
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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悠闲数学娱乐论坛(第3版)
2025-3-5 12:22 GMT+8
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