Question
Using principle of mathematical induction prove that $\sqrt{\text{n}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{n}}}$ for all natural numbers $\text{n}\geq2.$

Answer

 Let p(n) be the statement given by

P(n):$\sqrt{\text{n}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{n}}}$ for all natural numbers $\text{n}\geq2.$

Step 1:

p(2): $\sqrt{2}=1.4142$

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1+\frac{1}{1.4142}=1+0.7071=1.7071$

$\therefore\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$

$\therefore$ p(2) is true.

Step 2:

Let p(m) is true. Then,

$\sqrt{\text{m}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}} \ ...(1)$

We have to prove that p(m + 1) is true.

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}>\sqrt{\text{m}} \ ...[\text{From}(1)]$

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\sqrt{\text{m}}+\frac{1}{\sqrt{\text{m}+1}}$

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\frac{\sqrt{\text{m}^2+\text{m}}+1}{\sqrt{\text{m}+1}}$

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\frac{\sqrt{\text{m}^2}+1}{\sqrt{\text{m}+1}}$

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\frac{\text{m}+1}{\sqrt{\text{m}+1}}$

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\sqrt{\text{m}+1}$

⇒ p(m + 1) is true.

Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$ 

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