# √n < 1/√1 + 1/√2 + … 1/√n,

Question:

√n < 1/√1 + 1/√2 + … 1/√n, for all natural numbers n ≥ 2.

Solution:

According to the question,

$P(n)$ is $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots \ldots \ldots . .+\frac{1}{\sqrt{n}} ; n \geq 2 P(n)$ is $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots \ldots \ldots . .+$

$\frac{1}{\sqrt{n}} ; n \geq 2$

$\mathrm{P}(2)$ is $\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}} \Rightarrow 1.414<1.707$ It's true

$\mathrm{P}(3)$ is $\sqrt{3}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \Rightarrow 1.732<2.284 \mathrm{It}^{\prime} \mathrm{s}$ true

Let $P(k)=\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots \ldots \ldots . .+\frac{1}{\sqrt{k}} ;$ is true

Adding $\sqrt{\mathrm{k}+1}-\sqrt{\mathrm{k}}$ on both sides.

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

$\left[\because \sqrt{\mathrm{k}+1}-\sqrt{\mathrm{k}}=\frac{(\sqrt{\mathrm{k}+1}-\sqrt{\mathrm{k}})(\sqrt{\mathrm{k}+1}+\sqrt{\mathrm{k}})}{(\sqrt{\mathrm{k}+1}+\sqrt{\mathrm{k}})}=\frac{1}{(\sqrt{\mathrm{k}+1}+\sqrt{\mathrm{k}})} \leq \frac{1}{\sqrt{\mathrm{k}+1}}\right]$

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

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

√n < 1/√1 + 1/√2 + … 1/√n, for all natural numbers n ≥ 2