# Can you help me?

• June 24th 2006, 11:38 AM
Nichelle14
Can you help me?
Prove that 3^n > n^3 for all positive integers n.

I first tried n= 1, which showed 3>1
n= 2, 9>8
but n = 3 became 27 = 27, which is not true, but everything else works afterwards.

I was given the hint to use induction. So I tried this:

3^(n+1) > (n+1)^3
3*(3^n) > n^3 + 3n^2 + 3n + 1

What do I do next?
• June 24th 2006, 01:35 PM
Soltras
Suppose $3^n > n^3$ for some $n = k, k \ge 4$ which was shown.

For the inductive step, you want to show that $3^{k+1} > (k+1)^3$.

Well $(k+1)^3 = k^3 + 3k^2 + 3k + 1$.

Since it is known that $k \ge 4$ then $k^3 \ge 4k^2 > 3k^2$.

Also, $k \ge 4$ leads to $k^2 \ge 16$, so $k^2 - 3 \ge 13$ and it follows that $k(k^2-3) \ge 52 > 1$. Rearranging, it is evident that $3k + 1 < k^3$.

So applying the above revelations, $k^3 + 3k^2 + 3k + 1 = (k^3) + (3k^2) + (3k+1)$
$< k^3 + k^3 + k^3 = 3k^3.$

It is hypothesized that $k^3 < 3^k$, so $3k^3 < 3 \cdot 3^k = 3^{k+1}.$

Thus, $(k+1)^3 < 3^{k+1}$, and by induction, $3^n > n^3$ is true for all $n \ge 4$.
• June 24th 2006, 01:44 PM
Nichelle14
Thank you for your help. But I need a little more explantion.

I don't quite know where or why you did this particular line: k^3 >= 4k^2>3k^2

Then why did you do the two lines after the above line?
• June 24th 2006, 02:15 PM
Soltras
I'll try to be clearer:
I wanted to make it clear that $k^3 > 3k^2$ and $k^3 > 3k+1$.

Using both of these facts, then $k^3+k^3 > 3k^2 + 3k +1$, so $k^3+k^3+k^3 > k^3 + 3k^2 + 3k +1$ which is the expansion of $(k+1)^3$.

So all of the above sets up:

$(k+1)^3 < 3k^3$

and since $k^3 < 3^k$, then $3k^3 < 3 \cdot 3^k$. in other words, $(k+1)^3 < 3^{k+1}$, which is what we wre trying to show.