Proving inequality by mathematical induction

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Homework Help Overview

The problem involves proving the inequality 2n < (n+1)! for n≥2 using mathematical induction. Participants are discussing the steps involved in the proof and the validity of the reasoning presented.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants describe the base case verification for n=2 and the induction hypothesis. There is discussion on how to manipulate the inequality and the implications of multiplying both sides by 2. Some participants question the assumptions made regarding the factorial growth compared to the exponential growth of 2n.

Discussion Status

Some participants express confidence in the understanding of induction proofs, while others seek clarification on specific reasoning steps. There is no explicit consensus on the final validity of the proof, but the discussion is productive and focused on refining the arguments presented.

Contextual Notes

Participants note that the problem was presented as an extra credit challenge, which may influence the level of scrutiny and attention to detail in the discussion.

dustbin
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Homework Statement



I am asked to prove:
2n < (n+1)! , where n≥2

The Attempt at a Solution



Base step: set n=2, then test 22 < (2+1)!

22 = 4
(2+1)!= 3! = 3(2)(1) = 6
so 4 < 6 , which is true.

Induction hypothesis is 2k < (k+1)!
Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)!

Attempt to solve:

starting with what I know: 2k < (k+1)!
Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)!

I know that 2(k+1)! < (k+2)!
since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.

Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!.

P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.Thanks for any help!
EDIT: fixed a couple of type-o's.
 
Last edited:
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dustbin said:

Homework Statement



I am asked to prove:
2n < (n+1)! , where n≥2

The Attempt at a Solution



Base step: set n=2, then test 22 < (2+1)!

22 = 4
(2+1)!= 3! = 3(2)(1) = 6
so 4 < 6 , which is true.

Induction hypothesis is 2k < (k+1)!
Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)!

Attempt to solve:

starting with what I know: 2k < (k+1)!
Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)!

I know that 2(k+1)! < (k+2)!
since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.

Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!.

P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.Thanks for any help!
EDIT: fixed a couple of type-o's.

Welcome to PF, and very nice first post! Looks good!
 
Thank you very much for your response! It was posted as an extra credit problem by my professor. I wanted to make sure my reasoning was correct before posting it on the board, as I've been having a little difficulty grasping what we covered about mathematical induction.
 
It looks to me like you have the idea of induction proofs.
 

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