\(S=\frac{2}{2}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{\sqrt{100}}\)

\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{100}+\sqrt{101}}\)

\(S>2\left(\sqrt{2}-\sqrt{1}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{101}-\sqrt{100}\right)\)

\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\) (1)

\(S< 1+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

\(S< 1+2\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{100}-\sqrt{99}\right)\)

\(S< 1+2\left(\sqrt{100}-1\right)=19\) (2)

(1); (2) \(\Rightarrow18< S< 19\)

a

\(\frac{1}{\sqrt{n}}=\frac{2}{2\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(\Rightarrow S>1+2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\right)\)

\(\Rightarrow S>1+2\left(\sqrt{n+1}-\sqrt{2}\right)\)

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

Với n = 2 thì \(\frac{1}{1}+\frac{1}{\sqrt{2}}>\sqrt{2}\)

Giả sử bất đẳng thức đúng đến n = k

=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{K}}>\sqrt{K}\)

Ta chứng minh bất đẳng thức đúng với n = k+1

Ta có \(\frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{K}}+\frac{1}{\sqrt{K+1}}>\sqrt{K}+\frac{1}{\sqrt{K+1}}\)

= \(\frac{1+\sqrt{K^2+K}}{\sqrt{K+1}}\)

Mà ta lại có

\(\frac{1+\sqrt{K^2+K}}{\sqrt{K+1}}-\sqrt{K+1}\)

= \(\frac{\sqrt{K^2+K}-K}{\sqrt{K+1}}>0\)

Vậy bất đẳng thức đúng với n = k + 1

=> Điều phải chứng minh

Ta có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{n}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{n}};\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{n}};...\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}.n=\sqrt{n}\)

\(S=\frac{2}{2}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{100}}\)

\(S< 1+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

\(S< 1+2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(S< 1+2\left(\sqrt{100}-\sqrt{1}\right)=19\)

\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{100}+\sqrt{101}}\)

\(S>2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)

\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\)

\(\Rightarrow18< S< 19\Rightarrow S\) ko là số tự nhiên