Question

chứng minh \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 18\)

Answer 1

Ta có: \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}=2.\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\) (1)

\(\left(1\right)< 2.\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\right)\)\(=2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)\(=2\left(-\sqrt{1}+\sqrt{100}\right)=2\left(-1+10\right)=18\)

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