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?
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
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 $\displaystyle 2n^3>6n^2,\;\;\;n>3$