\(\frac{1}{\sqrt{n}}=\frac{2}{2\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(\Rightarrow S>1+2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\right)\)
\(\Rightarrow S>1+2\left(\sqrt{n+1}-\sqrt{2}\right)\)
Với \(n\ge98\Rightarrow S>1+2\left(\sqrt{99}-\sqrt{2}\right)\)
Ta sẽ chứng minh \(1+2\left(\sqrt{99}-\sqrt{2}\right)>18\Leftrightarrow\sqrt{99}-\sqrt{2}>\frac{17}{2}\)
\(\Leftrightarrow101-2\sqrt{198}>\frac{298}{4}\Leftrightarrow\sqrt{198}< \frac{115}{8}\)
\(\Leftrightarrow198< \frac{13225}{64}\) (đúng vì \(\frac{13225}{64}>\frac{12800}{64}=200>198\))
Khi \(n=98\Rightarrow S>18\) theo cmt
Mặt khác: \(\frac{1}{\sqrt{n}}=\frac{2}{2\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\)
\(\Rightarrow S< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{98}-\sqrt{97}\right)\)
\(\Rightarrow S< 1+2\left(\sqrt{98}-1\right)=2\sqrt{98}-1< 2\sqrt{100}-1=19\)
\(\Rightarrow18< S< 19\Rightarrow S\) nằm giữa 2 STN liên tiếp nên ko thể là STN
