Giải thích các bước giải:
Ta có:
$A=1+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{n}}$
$\to \dfrac12A=\dfrac12+\dfrac{1}{2\sqrt{2}}+....+\dfrac{1}{2\sqrt{n}}$
$\to \dfrac12A=\dfrac1{2}+\dfrac{1}{\sqrt{2}+\sqrt{2}}+....+\dfrac{1}{\sqrt{n}+\sqrt{n}}$
$\to \dfrac12A<\dfrac12+\dfrac{1}{\sqrt{1}+\sqrt{2}}+....+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}$
$\to \dfrac12A<\dfrac12+\dfrac{1}{\sqrt{2}+\sqrt{1}}+....+\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$
$\to \dfrac12A<\dfrac12+\dfrac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}+....+\dfrac{\sqrt{n}-\sqrt{n-1}}{(\sqrt{n}+\sqrt{n-1})(\sqrt{n}-\sqrt{n-1})}$
$\to \dfrac12A<\dfrac12+\dfrac{\sqrt{2}-\sqrt{1}}{2-1}+....+\dfrac{\sqrt{n}-\sqrt{n-1}}{n-(n-1)}$
$\to \dfrac12A<\dfrac12+\dfrac{\sqrt{2}-\sqrt{1}}{1}+....+\dfrac{\sqrt{n}-\sqrt{n-1}}{1}$
$\to \dfrac12A<\dfrac12+(\sqrt{2}-\sqrt{1})+....+(\sqrt{n}-\sqrt{n-1})$
$\to \dfrac12A<\dfrac12-\sqrt{1}+\sqrt{2}-....-\sqrt{n-1}+\sqrt{n}$
$\to \dfrac12A<\dfrac12-1+\sqrt{2}-....-\sqrt{n-1}+\sqrt{n}$
$\to \dfrac12A<\sqrt{n}-\dfrac12$
$\to \dfrac12A<\sqrt{n}$
$\to A<2\sqrt{n}$
