Using principle of mathematical induction, prove that

Question:

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

Solution:

Let $\mathrm{P}(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.

Step I: For $n=2$,

$\mathrm{P}(2)$ :

LHS $=\sqrt{2} \approx 1.414$

$\mathrm{RHS}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1+\frac{\sqrt{2}}{2} \approx 1+0.707=1.707$

As, LHS $<$ RHS

So, it is true for $n=2$.

Step II : For $n=k$,

Let $\mathrm{P}(k): \sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{k}}$ be true for some natural numbers $n \geq 2$.

Step III : For $n=k+1$,

$\mathrm{P}(k+1):$

$\mathrm{LHS}=\sqrt{k+1}$

$\mathrm{RHS}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}$

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

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

So, it is also true for $n=k+1$.

Hence, $\mathrm{P}(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.