cho A= $frac{1}{2^{2}}$ + $frac{1}{4^{2}}$ + $frac{1}{6^{2}}$ + ... + $frac{1}{100^{2}}$. Chứng tỏ A < $frac{1}{2}

2025-05-05 11:56:28

cho A= $\frac{1}{2^{2}}$ + $\frac{1}{4^{2}}$ + $\frac{1}{6^{2}}$ + ... + $\frac{1}{100^{2}}$. Chứng tỏ A < $\frac{1}{2}

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