Question

[Ngoc An Pham]

Chứng minh rằng:

\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{225}}< 28\)

Answer 1

[Nguyễn Xuân Tiến 24]

Đặt \(A=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{225}}\)

\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+...+\dfrac{2}{\sqrt{225}+\sqrt{225}}\)

\(\Rightarrow A< \dfrac{2}{\sqrt{2}+\sqrt{1}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}+...+\dfrac{2}{\sqrt{225}+\sqrt{224}}=\)

\(=2[\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+(\sqrt{225}-\sqrt{224})]\)

\(\Leftrightarrow A< 2.\left(\sqrt{225}-1\right)=2.14=28\left(đpcm\right)\)

Bài toán tổng quát:Chứng minh BĐT sau với \(n\in N;n\ge2\)

\(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)

Answer 2

[Ngoc An Pham]

thanks