Lời giải:

Liên hợp ta thấy:

\(2(\sqrt{n+1}-\sqrt{n})=2.\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(1)\)

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

Từ \((1);(2)\Rightarrow 2(\sqrt{n+1}-\sqrt{n})< \frac{1}{\sqrt{n}}< 2(\sqrt{n}-\sqrt{n-1})\)

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Áp dụng vào bài toán:

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

\(\Leftrightarrow S>1+2(\sqrt{101}-\sqrt{2})>18(*)\)

Và:

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

\(\Leftrightarrow S< 1+2(\sqrt{100}-\sqrt{1})=19(**)\)

Từ $(*); (**)$ suy ra $18< S< 19$ (đpcm)

Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)

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

Áp dụng vào bài toán ta có: 

\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

.....

\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)

Lời giải:

Xét số hạng tổng quát:

\(\frac{1}{(n+1)\sqrt{n+1}+n\sqrt{n}}=\frac{1}{(\sqrt{n}+\sqrt{n+1})[n+\sqrt{n(n+1)}+n+1)]}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n}+\sqrt{n+1})[n+\sqrt{n(n+1)}+n+1)]}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{n+(n+1)-\sqrt{n(n+1)}}<\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}-\sqrt{n(n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Do đó:

\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

......

\(\frac{1}{(n+1)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Cộng theo vế:

\(\Rightarrow \text{VT}< 1-\frac{1}{\sqrt{n+1}}\) (đpcm)

Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) (pp trục căn thức ở mẫu)

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

Áp dụng tính: \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)

                        \(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

                          \(=1-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

Vậy S = 19/20

 Xét số hạng tổng quát ta có:

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

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

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

Áp dụng vào bài tập, ta có:

\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)

\(< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+...+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)

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