Question

$\mathrm{Using}\mathrm{principle}\mathrm{of}\mathrm{mathematical}\mathrm{induction},\mathrm{prove}\mathrm{that}\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}n\ge 2.$ [NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{P}\left(n\right):\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}n\ge 2. \\ \mathrm{Step}\mathrm{I}:\mathrm{For}n=2, \\ \mathrm{P}\left(2\right): \\ \mathrm{LHS}=\sqrt{2}\approx 1.414 \\ \mathrm{RHS}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1+\frac{\sqrt{2}}{2}\approx 1+0.707=1.707 \\ \mathrm{As},\mathrm{LHS}<\mathrm{RHS} \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=2. \\ \mathrm{Step}\mathrm{II}:\mathrm{For}n=k, \\ \mathrm{Let}\mathrm{P}\left(k\right):\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{k}}\mathrm{be}\mathrm{true}\mathrm{for}\mathrm{some}\mathrm{natural}\mathrm{numbers}n\ge 2. \\ \mathrm{Step}\mathrm{III}:\mathrm{For}n=k+1, \\ \mathrm{P}\left(k+1\right): \\ \mathrm{LHS}=\sqrt{k+1} \\ \mathrm{RHS}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \\ >\sqrt{k}+\frac{1}{\sqrt{k+1}} \\ \mathrm{As},\sqrt{k+1}>\sqrt{k} \\ \Rightarrow \frac{\sqrt{k}}{\sqrt{k+1}}<1 \\ \Rightarrow \frac{k}{\sqrt{k+1}}<\sqrt{k} \\ \Rightarrow \frac{k+1}{\sqrt{k+1}}-\frac{1}{\sqrt{k+1}}<\sqrt{k} \\ \Rightarrow \sqrt{k+1}-\frac{1}{\sqrt{k+1}}<\sqrt{k} \\ \Rightarrow \sqrt{k}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1} \\ \mathrm{i}.\mathrm{e}.\mathrm{LHS}<\mathrm{RHS} \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{also}\mathrm{true}\mathrm{for}n=k+1. \\ \mathrm{Hence},\mathrm{P}\left(n\right):\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}n\ge 2. \end{gathered}$$