Chứng minh  $\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\mathrm{...}+\frac{1}{2005\sqrt{2004}}<2$.

Với mọi n ∈ ℕ* ta có

 $\iff {\left(\sqrt{n+1}-\sqrt{n}\right)}^2>0$ (luôn đúng)

Áp dụng bất đẳng thức trên ta có

...

Suy ra

$\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\mathrm{...}+\frac{1}{2005\sqrt{2004}}<2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\mathrm{...}+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)$

$\iff \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\mathrm{...}+\frac{1}{2005\sqrt{2004}}<2\left(1-\frac{1}{\sqrt{2005}}\right)<2$

Vậy  $\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\mathrm{...}+\frac{1}{2005\sqrt{2004}}<2$.