Prove the statement by the Principle of Mathematical Induction: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.
(N/A) Let $P(n): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}}$ for $n \geq 2$.
For $n=2$,$P(2): \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{1.414} \approx 1 + 0.707 = 1.707$. Since $\sqrt{2} \approx 1.414$,$1.414 < 1.707$ is true.
Assume $P(k)$ is true for some $k \geq 2$: $\sqrt{k} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}}$.
We need to show $P(k+1): \sqrt{k+1} < \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
We know $\sqrt{k+1} - \sqrt{k} = \frac{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1} + \sqrt{k})}{\sqrt{k+1} + \sqrt{k}} = \frac{1}{\sqrt{k+1} + \sqrt{k}}$.
Since $\sqrt{k+1} + \sqrt{k} > 2\sqrt{k} > \sqrt{k+1}$,we have $\frac{1}{\sqrt{k+1} + \sqrt{k}} < \frac{1}{\sqrt{k+1}}$.
Thus,$\sqrt{k+1} < \sqrt{k} + \frac{1}{\sqrt{k+1}} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{k+1}} = \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.
For $n=2$,$P(2): \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{1.414} \approx 1 + 0.707 = 1.707$. Since $\sqrt{2} \approx 1.414$,$1.414 < 1.707$ is true.
Assume $P(k)$ is true for some $k \geq 2$: $\sqrt{k} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}}$.
We need to show $P(k+1): \sqrt{k+1} < \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
We know $\sqrt{k+1} - \sqrt{k} = \frac{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1} + \sqrt{k})}{\sqrt{k+1} + \sqrt{k}} = \frac{1}{\sqrt{k+1} + \sqrt{k}}$.
Since $\sqrt{k+1} + \sqrt{k} > 2\sqrt{k} > \sqrt{k+1}$,we have $\frac{1}{\sqrt{k+1} + \sqrt{k}} < \frac{1}{\sqrt{k+1}}$.
Thus,$\sqrt{k+1} < \sqrt{k} + \frac{1}{\sqrt{k+1}} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{k+1}} = \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.
Explore More
Similar Questions
Prove the following by using the principle of mathematical induction for all $n \in N:$
$3^{2n} - 1$ is divisible by $8$.
$3^{2n} - 1$ is divisible by $8$.
Difficult
View SolutionIf $P(n) = 2 + 4 + 6 + \dots + 2n$,$n \in N$,then $P(k) = k(k + 1) + 2 \implies P(k + 1) = (k + 1)(k + 2) + 2$ for all $k \in N$. So we can conclude that $P(k) = k(k + 1) + 2$ for all $k \in N$ is true. What can we conclude about $P(n) = n(n + 1) + 2$ for all $n \in N$?
Easy
View SolutionProve that for all $n \in N$,$3^{2n+2} - 8n - 9$ is divisible by $8$ using the principle of mathematical induction.
Difficult
View SolutionProve the statement by the Principle of Mathematical Induction: $n^{2} < 2^{n}$ for all natural numbers $n \geq 5$.
Difficult
View SolutionLet $S(k) = 1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$. Then which of the following is true?
MediumAIEEE 2004
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo