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

Question

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.

Answer 1

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

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

Answer 2

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

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

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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.

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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.

Answer 3

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! $

Answer 4

Short answer:

$$4!>3\cdot4$$

$$n!>3n\implies(n+1)!=(n+1)n!>(n+1)3n>3(n+1).$$

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

$$1\cdot2\cdot\color{green}3\cdot4\cdots\color{green}n>\color{green}{3n}.$$

Answer 5

No need for induction ! For $n \ge 4$

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