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Proving that $n! > 3n$ for $n geq 4$ by induction

Proof that $n! > 3n$ for $nge4 $ using the Principle of Mathematical InductionProof that $n! > 3n$ for $nge4 $ using the Principle of Mathematical InductionProper way to make transition from $k$ to $k+1$ in proofs by inductionthe purpose of inductionProving $n! > n^2$ by mathematical inductionHow to prove this inequality $3^{n}geq n^{2}$ for $ngeq 1$ with mathematical induction?How to prove inequality $3^{n}≤4n!$ for $n≥4$ with mathematical induction?Recursive induction for a sequence.Prove by induction that for $ngeq1$, $(a_1+a_2+…+a_n)/n geq (a_1*a_2*…*a_n)^{1/n}$Prove by Induction that $2^n geq n+12$ for all $ngeq4$Explaining the proof of Fibonacci number using inductive reasoning

2

$begingroup$

I need to prove that $n! > 3n$ for $n geq 4$ by induction. There are countless explanations of how to do it for say $3^n$ but the only relevant example I can find doesn't give me a clear answer of how to do it.

When doing the inductive step I get to

$$(n+1)n! > (n+1)3n$$

but not sure what to do after that.

logic induction

share|cite|improve this question

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

asked Mar 12 at 5:17

Sam WoodSam Wood

132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is that what you've deduced or what you want to prove?
  $endgroup$
  – saulspatz
  Mar 12 at 5:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't know how to prove $(n+1)3nge 3(n+1) $???
  $endgroup$
  – fleablood
  Mar 12 at 6:00
  

  
  
  
  

add a comment | 

2

$begingroup$

I need to prove that $n! > 3n$ for $n geq 4$ by induction. There are countless explanations of how to do it for say $3^n$ but the only relevant example I can find doesn't give me a clear answer of how to do it.

When doing the inductive step I get to

$$(n+1)n! > (n+1)3n$$

but not sure what to do after that.

logic induction

share|cite|improve this question

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

asked Mar 12 at 5:17

Sam WoodSam Wood

132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is that what you've deduced or what you want to prove?
  $endgroup$
  – saulspatz
  Mar 12 at 5:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't know how to prove $(n+1)3nge 3(n+1) $???
  $endgroup$
  – fleablood
  Mar 12 at 6:00
  

  
  
  
  

add a comment | 

$begingroup$

I need to prove that $n! > 3n$ for $n geq 4$ by induction. There are countless explanations of how to do it for say $3^n$ but the only relevant example I can find doesn't give me a clear answer of how to do it.

When doing the inductive step I get to

$$(n+1)n! > (n+1)3n$$

but not sure what to do after that.

logic induction

share|cite|improve this question

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

asked Mar 12 at 5:17

Sam WoodSam Wood

132

$endgroup$

share|cite|improve this question

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

asked Mar 12 at 5:17

Sam WoodSam Wood

132

share|cite|improve this question

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

asked Mar 12 at 5:17

Sam WoodSam Wood

132

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

edited Mar 12 at 5:27

Eevee Trainer

8,07421439

asked Mar 12 at 5:17

Sam WoodSam Wood

132

asked Mar 12 at 5:17

Sam WoodSam Wood

132

5 Answers
5

active

oldest

votes

1

$begingroup$

Since $(n+1)3n>(n+1)3$ you don't have to go any further. You are done.

Induction Step:

Assume $n!>3n$. Then

$(n+1)! =$

$(n+1)n!> $

$(n+1)3n> $

$3(n+1)$

So $n!>3nimplies (n+1)!>3 (n+1) $

That's all. (Assuming you did the base case: $4!>3*4$.)

share|cite|improve this answer

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, i was confusing myself and thinking that i had to do more to prove it.
  $endgroup$
  – Sam Wood
  Mar 12 at 6:57
  

  
  
  
  

add a comment | 

0

$begingroup$

Option.

$n ge 4.$

0) Base case n=4√

1)Hypothesis: $3n < n!$

Step $n+1$:

$3(n+1) =3n +3 < n! +3 <$

$n! +nn!= (n+1)n!=(n+1)!.$

Used: $3 < nn! $

share|cite|improve this answer

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

add a comment | 

0

$begingroup$

Short answer:

$$4!>3cdot4$$

$$n!>3nimplies(n+1)!=(n+1)n!>(n+1)3n>3(n+1).$$

---

Anyway, this theorem really doesn't deserve an inductive proof. It is immediate that

$$1cdot2cdotcolor{green}3cdot4cdotscolor{green}n>color{green}{3n}.$$

share|cite|improve this answer

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

0

$begingroup$

No need for induction ! For $n ge 4$

$frac{n!}{3n} = frac{(n-1)!}{3} ge 2\ Rightarrow n! ge 6n > 3n$

share|cite|improve this answer

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

$endgroup$

add a comment | 

0

$begingroup$

For your proof, you need to have three things:

• 
  A base case: Verify the identity for your "lowest/base" value. Here, that is $n=4$.


• 
  The inductive hypothesis: This is what you assume holds. For some $n=k>4$, here, we will assume this holds:

$$k! > 3k$$

• 
  The inductive step: Now that we assume the proposition holds for $n=k$, we want to show that the "next highest" value also holds. Here, that means we want to show $n=k+1$ holds. That means we want to show

$$(k+1)! > 3(k+1)$$

---

You seem to have had some sort of misunderstanding on the last step because what you're trying to prove isn't right there. Start with what I have above and see what you can do.

Some further tips:

• You will want to note that the inductive hypothesis is very important in inductive proofs, in that you will want to use it in the induction step at some point. Try to manipulate the expression in question to make the hypothesis pop out.*




• An occasionally overlooked property of the factorial is that $(n+1)! = n! cdot (n+1)$. This may prove useful in this proof.

---

Footnote (*): I don't really think you'll need to make direct use of the inductive hypothesis here, at least the way I would do this proof by induction. I'd also argue that one doesn't need induction to prove this in the first place but I imagine this is mostly just a contrived example to help you learn induction.

share|cite|improve this answer

edited Mar 15 at 1:45

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a solution, but an extended comment.
  $endgroup$
  – DavidG
  Mar 15 at 1:42
  

  
  
  
  

add a comment | 

Your Answer

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1

$begingroup$

Since $(n+1)3n>(n+1)3$ you don't have to go any further. You are done.

Induction Step:

Assume $n!>3n$. Then

$(n+1)! =$

$(n+1)n!> $

$(n+1)3n> $

$3(n+1)$

So $n!>3nimplies (n+1)!>3 (n+1) $

That's all. (Assuming you did the base case: $4!>3*4$.)

share|cite|improve this answer

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, i was confusing myself and thinking that i had to do more to prove it.
  $endgroup$
  – Sam Wood
  Mar 12 at 6:57
  

  
  
  
  

add a comment | 

1

$begingroup$

Since $(n+1)3n>(n+1)3$ you don't have to go any further. You are done.

Induction Step:

Assume $n!>3n$. Then

$(n+1)! =$

$(n+1)n!> $

$(n+1)3n> $

$3(n+1)$

So $n!>3nimplies (n+1)!>3 (n+1) $

That's all. (Assuming you did the base case: $4!>3*4$.)

share|cite|improve this answer

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, i was confusing myself and thinking that i had to do more to prove it.
  $endgroup$
  – Sam Wood
  Mar 12 at 6:57
  

  
  
  
  

add a comment | 

$begingroup$

Since $(n+1)3n>(n+1)3$ you don't have to go any further. You are done.

Induction Step:

Assume $n!>3n$. Then

$(n+1)! =$

$(n+1)n!> $

$(n+1)3n> $

$3(n+1)$

So $n!>3nimplies (n+1)!>3 (n+1) $

That's all. (Assuming you did the base case: $4!>3*4$.)

share|cite|improve this answer

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

$endgroup$

share|cite|improve this answer

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

answered Mar 12 at 6:05

fleabloodfleablood

72.8k22788

0

$begingroup$

Option.

$n ge 4.$

0) Base case n=4√

1)Hypothesis: $3n < n!$

Step $n+1$:

$3(n+1) =3n +3 < n! +3 <$

$n! +nn!= (n+1)n!=(n+1)!.$

Used: $3 < nn! $

share|cite|improve this answer

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

add a comment | 

0

$begingroup$

Option.

$n ge 4.$

0) Base case n=4√

1)Hypothesis: $3n < n!$

Step $n+1$:

$3(n+1) =3n +3 < n! +3 <$

$n! +nn!= (n+1)n!=(n+1)!.$

Used: $3 < nn! $

share|cite|improve this answer

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

add a comment | 

$begingroup$

Option.

$n ge 4.$

0) Base case n=4√

1)Hypothesis: $3n < n!$

Step $n+1$:

$3(n+1) =3n +3 < n! +3 <$

$n! +nn!= (n+1)n!=(n+1)!.$

Used: $3 < nn! $

share|cite|improve this answer

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

answered Mar 12 at 9:16

Peter SzilasPeter Szilas

11.6k2822

0

$begingroup$

Short answer:

$$4!>3cdot4$$

$$n!>3nimplies(n+1)!=(n+1)n!>(n+1)3n>3(n+1).$$

---

Anyway, this theorem really doesn't deserve an inductive proof. It is immediate that

$$1cdot2cdotcolor{green}3cdot4cdotscolor{green}n>color{green}{3n}.$$

share|cite|improve this answer

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

0

$begingroup$

Short answer:

$$4!>3cdot4$$

$$n!>3nimplies(n+1)!=(n+1)n!>(n+1)3n>3(n+1).$$

---

Anyway, this theorem really doesn't deserve an inductive proof. It is immediate that

$$1cdot2cdotcolor{green}3cdot4cdotscolor{green}n>color{green}{3n}.$$

share|cite|improve this answer

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

$begingroup$

Short answer:

$$4!>3cdot4$$

$$n!>3nimplies(n+1)!=(n+1)n!>(n+1)3n>3(n+1).$$

---

Anyway, this theorem really doesn't deserve an inductive proof. It is immediate that

$$1cdot2cdotcolor{green}3cdot4cdotscolor{green}n>color{green}{3n}.$$

share|cite|improve this answer

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

answered Mar 12 at 9:23

Yves DaoustYves Daoust

131k676229

0

$begingroup$

No need for induction ! For $n ge 4$

$frac{n!}{3n} = frac{(n-1)!}{3} ge 2\ Rightarrow n! ge 6n > 3n$

share|cite|improve this answer

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

$endgroup$

add a comment | 

0

$begingroup$

No need for induction ! For $n ge 4$

$frac{n!}{3n} = frac{(n-1)!}{3} ge 2\ Rightarrow n! ge 6n > 3n$

share|cite|improve this answer

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

$endgroup$

add a comment | 

$begingroup$

No need for induction ! For $n ge 4$

$frac{n!}{3n} = frac{(n-1)!}{3} ge 2\ Rightarrow n! ge 6n > 3n$

share|cite|improve this answer

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

answered Mar 12 at 9:31

gandalf61gandalf61

9,109825

0

$begingroup$

For your proof, you need to have three things:

• 
  A base case: Verify the identity for your "lowest/base" value. Here, that is $n=4$.


• 
  The inductive hypothesis: This is what you assume holds. For some $n=k>4$, here, we will assume this holds:

$$k! > 3k$$

• 
  The inductive step: Now that we assume the proposition holds for $n=k$, we want to show that the "next highest" value also holds. Here, that means we want to show $n=k+1$ holds. That means we want to show

$$(k+1)! > 3(k+1)$$

---

You seem to have had some sort of misunderstanding on the last step because what you're trying to prove isn't right there. Start with what I have above and see what you can do.

Some further tips:

• You will want to note that the inductive hypothesis is very important in inductive proofs, in that you will want to use it in the induction step at some point. Try to manipulate the expression in question to make the hypothesis pop out.*




• An occasionally overlooked property of the factorial is that $(n+1)! = n! cdot (n+1)$. This may prove useful in this proof.

---

Footnote (*): I don't really think you'll need to make direct use of the inductive hypothesis here, at least the way I would do this proof by induction. I'd also argue that one doesn't need induction to prove this in the first place but I imagine this is mostly just a contrived example to help you learn induction.

share|cite|improve this answer

edited Mar 15 at 1:45

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a solution, but an extended comment.
  $endgroup$
  – DavidG
  Mar 15 at 1:42
  

  
  
  
  

add a comment | 

0

$begingroup$

For your proof, you need to have three things:

• 
  A base case: Verify the identity for your "lowest/base" value. Here, that is $n=4$.


• 
  The inductive hypothesis: This is what you assume holds. For some $n=k>4$, here, we will assume this holds:

$$k! > 3k$$

• 
  The inductive step: Now that we assume the proposition holds for $n=k$, we want to show that the "next highest" value also holds. Here, that means we want to show $n=k+1$ holds. That means we want to show

$$(k+1)! > 3(k+1)$$

---

You seem to have had some sort of misunderstanding on the last step because what you're trying to prove isn't right there. Start with what I have above and see what you can do.

Some further tips:

• You will want to note that the inductive hypothesis is very important in inductive proofs, in that you will want to use it in the induction step at some point. Try to manipulate the expression in question to make the hypothesis pop out.*




• An occasionally overlooked property of the factorial is that $(n+1)! = n! cdot (n+1)$. This may prove useful in this proof.

---

Footnote (*): I don't really think you'll need to make direct use of the inductive hypothesis here, at least the way I would do this proof by induction. I'd also argue that one doesn't need induction to prove this in the first place but I imagine this is mostly just a contrived example to help you learn induction.

share|cite|improve this answer

edited Mar 15 at 1:45

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a solution, but an extended comment.
  $endgroup$
  – DavidG
  Mar 15 at 1:42
  

  
  
  
  

add a comment | 

$begingroup$

For your proof, you need to have three things:

• 
  A base case: Verify the identity for your "lowest/base" value. Here, that is $n=4$.


• 
  The inductive hypothesis: This is what you assume holds. For some $n=k>4$, here, we will assume this holds:

$$k! > 3k$$

• 
  The inductive step: Now that we assume the proposition holds for $n=k$, we want to show that the "next highest" value also holds. Here, that means we want to show $n=k+1$ holds. That means we want to show

$$(k+1)! > 3(k+1)$$

---

You seem to have had some sort of misunderstanding on the last step because what you're trying to prove isn't right there. Start with what I have above and see what you can do.

Some further tips:

• You will want to note that the inductive hypothesis is very important in inductive proofs, in that you will want to use it in the induction step at some point. Try to manipulate the expression in question to make the hypothesis pop out.*




• An occasionally overlooked property of the factorial is that $(n+1)! = n! cdot (n+1)$. This may prove useful in this proof.

---

Footnote (*): I don't really think you'll need to make direct use of the inductive hypothesis here, at least the way I would do this proof by induction. I'd also argue that one doesn't need induction to prove this in the first place but I imagine this is mostly just a contrived example to help you learn induction.

share|cite|improve this answer

edited Mar 15 at 1:45

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439

$endgroup$

share|cite|improve this answer

edited Mar 15 at 1:45

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439

answered Mar 12 at 5:26

Eevee TrainerEevee Trainer

8,07421439