Question

\[\text{ Using principle of mathematical induction, prove that } \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 .\]

 

Answer 1

\[\text { Let }  P\left( n \right): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 . \]
\[\text{ Step I: For }  n = 2, \]
\[P\left( 2 \right): \]
\[LHS = \sqrt{2} \approx 1 . 414\]
\[RHS = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{\sqrt{2}}{2} \approx 1 + 0 . 707 = 1 . 707\]
\[\text{ As, LHS < RHS } \]
\[\text{ So, it is true for }  n = 2 . \]
\[ \text{Step II : For } n = k \]
\[\text{ Let}  P\left( k \right): \sqrt{k} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{k}} \text{ be true for some natural numbers } n \geq 2 . \]
\[\text{ Step III: For } n = k + 1, \]
\[P\left( k + 1 \right): \]
\[LHS = \sqrt{k + 1}\]
\[RHS = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}}\]
\[ > \sqrt{k} + \frac{1}{\sqrt{k + 1}}\]
\[\text{ As, } \sqrt{k + 1} > \sqrt{k}\]
\[ \Rightarrow \frac{\sqrt{k}}{\sqrt{k + 1}} < 1\]
\[ \Rightarrow \frac{k}{\sqrt{k + 1}} < \sqrt{k}\]
\[ \Rightarrow \frac{k + 1}{\sqrt{k + 1}} - \frac{1}{\sqrt{k + 1}} < \sqrt{k}\]
\[ \Rightarrow \sqrt{k + 1} - \frac{1}{\sqrt{k + 1}} < \sqrt{k}\]
\[ \Rightarrow \sqrt{k} + \frac{1}{\sqrt{k + 1}} > \sqrt{k + 1}\]
\[i . e . LHS < RHS\]
\[\text{ So, it is also true for n }  = k + 1 . \]
\[\text{ Hence,}  P\left( n \right): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers }  n \geq 2 .\]