\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{225}}\)
\(\dfrac{1}{\sqrt{k}}=\dfrac{2}{\sqrt{k}+\sqrt{k}}< \dfrac{2}{\sqrt{k+1}+\sqrt{k}}\\ =\dfrac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)
\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{225}}\\ < 2\left(\sqrt{226}-\sqrt{225}\right)+2\left(\sqrt{225}-\sqrt{224}\right)+...+2\left(\sqrt{3}-\sqrt{2}\right)\\ =2\left(\sqrt{226}-\sqrt{225}+\sqrt{225}-\sqrt{224}+...+\sqrt{3}-\sqrt{2}\right)\\ =2\left(\sqrt{226}-\sqrt{2}\right)< 2\left(\sqrt{225}-\sqrt{2}\right)< 2\left(\sqrt{225}-\sqrt{1}\right)=28\left(đpcm\right)\)
Vậy \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{225}}< 28\)
