Name: Principle of Mathematical Induction Illustration
Uploaded: 2019-07-19
Duration: 4 min 48 s
Description: We proved the formula for the sum of first “n” natural numbers using the principle of mathematical induction. Is there any formula for the sum of fractions?...

Question

Using principle of mathematical induction, prove that for all

n \in N, n \geq 2, \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} .

Accepted Answer

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

For $n = 2, \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}$ which is true.

So, $P (2)$ is true.

Let $P (n)$ be true for some $n = k$ .

Then,

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

Adding $\frac{1}{\sqrt{k + 1}}$ on both sides, we get

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

Now,

$\sqrt{k + 1} < \sqrt{k} + \frac{1}{\sqrt{k + 1}} . . . . . (3)$

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

or $k < \sqrt{k^{2} + k}$ , which is true.

From $(2)$ and $(3)$

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

So, by the principle of mathematical induction, $P (n)$ is true for any natural number $n \geq 2$ .

Using principle of mathematical induction, prove that for all $n \in N, n \geq 2, \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} .$

Important Questions on Principle of Mathematical Induction

Learn and Prepare for any exam you want !

Using principle of mathematical induction, prove that for all n∈N, n≥2, n< 11+12+...+1n.

Important Questions on Principle of Mathematical Induction

Learn and Prepare for any exam you want !

Using principle of mathematical induction, prove that for all $n \in N, n \geq 2, \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} .$