Question

CMR : \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2018}}>2\left(\sqrt{2018}-1\right)\)

Answer 1

Đặt:   \(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}\)

Ta có:   \(\frac{1}{\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)       với \(\forall k\inℕ^∗\)

Do đó ta có:    \(A>2\left[\left(\sqrt{2019}-\sqrt{2018}\right)+\left(\sqrt{2018}-\sqrt{2017}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)\right]+1\)

  \(=2\left(\sqrt{2019}-\sqrt{2}\right)+1=2\sqrt{2019}-2\sqrt{2}+1>2\sqrt{2019}-3+1>2\sqrt{2019}-2\)

   \(>2\sqrt{2018}-2=2\left(\sqrt{2018}-1\right)\)

=>  đpcm