Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{4000}}$
$\to \dfrac12A=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+....+\dfrac{1}{2\sqrt{4000}}$
$\to \dfrac12A=\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+....+\dfrac{1}{\sqrt{4000}+\sqrt{4000}}$
$\to \dfrac12A<\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{3999}+\sqrt{4000}}$
$\to \dfrac12A<\dfrac{1}{\sqrt{4000}+\sqrt{3999}}+....+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{1}}$
$\to \dfrac12A<\dfrac{\sqrt{4000}-\sqrt{3999}}{(\sqrt{4000}+\sqrt{3999})(\sqrt{4000}-\sqrt{3999})}+....+\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\dfrac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}$
$\to \dfrac12A<\dfrac{\sqrt{4000}-\sqrt{3999}}{4000-3999}+....+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+\dfrac{\sqrt{2}-\sqrt{1}}{2-1}$
$\to \dfrac12A<\sqrt{4000}-\sqrt{3999}+....+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}$
$\to \dfrac12A<\sqrt{4000}-\sqrt{1}$
$\to \dfrac12A<19$
$\to A<38$
