For which integers is 3^n > n^3?

it seems to be for all but 3, but I am stuck on the proof.

Assume 3^n > n^3

n>3

Then

3^(n+1) > 3*n^3

Now what?

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- Aug 2nd 2011, 02:57 PMveronicak5678Inductive Proof
For which integers is 3^n > n^3?

it seems to be for all but 3, but I am stuck on the proof.

Assume 3^n > n^3

n>3

Then

3^(n+1) > 3*n^3

Now what? - Aug 2nd 2011, 04:14 PMPlatoRe: Inductive Proof
- Aug 2nd 2011, 04:49 PMArchie MeadeRe: Inductive Proof
Another way to consider...

You want to show

$\displaystyle 3^{n+1}>(n+1)^3$

Note that $\displaystyle 3^1>1^3,\;\;\;3^2>2^3,\;\;\;3^3=3^3$

Hence n>3 for the proof...

If $\displaystyle 3^n>n^3$, then $\displaystyle 3^{n+1}>3n^3$

If we can show that $\displaystyle 3n^3>(n+1)^3$, for n>3, the proof is complete.

Since we know n>3, then

$\displaystyle 3n^3>n^3+3n^2+3n+1\;\;?$

$\displaystyle 2n^3>3n^2+3n+1\;\;?$

Now use the fact that if n>3, then $\displaystyle 2n^3>2(3)n^2$ and continue - Aug 2nd 2011, 05:40 PMveronicak5678Re: Inductive Proof
Thanks for the answer, but I don't follow it.

What do you mean by the question marks?

We know n > 3, so 3n^3 > 3*81 - Aug 3rd 2011, 02:01 AMArchie MeadeRe: Inductive Proof
The purpose of the question marks is...

We are asking if $\displaystyle 3n^3>n^3+3n^2+3n+1,\;\;if\;\; n>3$

Because if that's true, then $\displaystyle 3^{n+1}>(n+1)^3$

Proof By Induction requires that you show $\displaystyle 3^{n+1}>(n+1)^3\;\;\;if\;\;\;3^n>n^3$

Hence we are asking if

$\displaystyle 2n^3\;\;is\;greater\;than\;\;\;3n^2+3n+1,\;\;n>3$

Now if n>3, then $\displaystyle 2n^3>2(3)n^2$

and you can finish by showing than $\displaystyle 6n^2>3n^2+3n+1$

using the same reasoning,

since $\displaystyle 2n^3>6n^2,\;\;\;n>3$ - Aug 3rd 2011, 03:02 AMSwlabrRe: Inductive Proof
- Aug 3rd 2011, 10:47 AMveronicak5678Re: Inductive Proof
Got it. Thanks very much, everyone!