Question

Prove by the Principle of Mathematical Induction: $\sqrt{n}<1/\sqrt{1}+1/\sqrt{2}+...+1/\sqrt{n},$ for all natural numbers $n\ge 2$.

Answer 1

Assume given statement

Let $P\left(n\right):\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{n},}\forall n\ge 2,nϵN$

Check that statement is true for $n=2$

$P\left(2\right):\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$

$\Rightarrow P\left(2\right):\sqrt{2}<\frac{\sqrt{2}+1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$\Rightarrow P\left(2\right):\sqrt{2}<\frac{2+\sqrt{2}}{2}\text{(TRUE)}$

So, P(n) is true for $n=2$

Assume P(k) to be true and then prove $P(k+1)$ is true.

Let P(n) is true for some natural number $n=k$

$\Rightarrow \sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{k}}\dots \left(1\right)$

To show: $\sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}+1}$

Now consider, $\sqrt{k}<\sqrt{k}+1$

$\Rightarrow \frac{1}{\sqrt{k}}>\frac{1}{\sqrt{k+1}}$

$\Rightarrow \frac{k}{\sqrt{k+1}}<\frac{k}{\sqrt{k}}$

$\Rightarrow \frac{k}{\sqrt{k+1}}<\sqrt{k}$

$\Rightarrow \frac{k+1}{\sqrt{k}+1}-\frac{1}{\sqrt{k}+1}<\sqrt{k}$

$\Rightarrow \sqrt{k}+1<\sqrt{k}+\frac{1}{\sqrt{k+1}}$

$\Rightarrow \sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\dots$

using (1)

Thus, $P(k+1)$is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers $n\ge 2$.