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thinhso01

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luffy_1998

Với $k \in Z, k \ge 2: \dfrac{1}{k^2} < \dfrac{1}{k(k-1)} = \dfrac{1}{k - 1} - \dfrac{1}{k}$
$\rightarrow 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + ... + \dfrac{1}{n^2} < 1 + 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{n^2-1} - \dfrac{1}{n} = 2 - \dfrac{1}{n}. (dpcm)$
 
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