Question

Using principle of mathematical induction prove that.

$\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}$ for all natural numbers $n\ge 2$.

Answer 1

Consider the statement

$P\left(n\right):\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}$, for all natural numbers n $\ge$ 2.

Step I We observe that P(2) is true

P(2) : $\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$, which is true,

Step II Now, assume that P(n) is true for n = k

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

Step III To prove P(k + 1) is true, we have to show that

$P(k+1):\sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{k+1}}$ is true.

Given that, $\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{k}}$

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

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

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

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

$\Rightarrow k<\sqrt{k(k+1)}\Rightarrow \sqrt{k}<\sqrt{k}+1$ .........(ii)

From eqs. (i) and (ii),

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

So, P(k + 1) is true, whenever P(k) is true.

Hence, P(n) is true.