Proving inequality by mathematical induction

AI Thread Summary
The discussion focuses on proving the inequality 2n < (n+1)! for n≥2 using mathematical induction. The base case is verified by testing n=2, confirming that 4 < 6. The induction hypothesis assumes 2k < (k+1)! and aims to prove that 2(k+1) < (k+2)!. By multiplying both sides of the hypothesis by 2 and recognizing that 2(k+1)! is less than (k+2)!, the induction step is completed. The conclusion is that the inequality holds true for all n≥2 by mathematical induction.
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.
 
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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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