# Thread: elem.num.theo - help w/proof

1. ## elem.num.theo - help w/proof

So - I know all the steps, I just need help with a gap.

Prove $n! > n^2 \exists n \geq4$
Ok, proof by Induction:

Since 4! = 24 which is $>(4)^2 = 16$, $n!>n^2$ for n=4.
Let k be a positive integer [math\geq4[/tex] and suppose $k!>k^2$.
Then, $k!(k+1) > k^2(k+1)$
$(k+1)! > k^2(k+1) = k^3+k^2$

Here's where I'm stuck. I know $k\geq4$ but how can I show that $(k+1)^2 < k^3+k^2$?

From then, I can show $(k+1)! > k^3+k^2 > (k+1)^2$
Therefore, $(k+1)! > (k+1)^2$, so $n! > n^2 for n \geq4$.

Please help me out with this way of solving the proof, I'm sure there are other ways.

2. Originally Posted by cassiopeia1289
So - I know all the steps, I just need help with a gap.

Prove $n! > n^2 \exists n \geq4$
Ok, proof by Induction:

Since 4! = 24 which is $>(4)^2 = 16$, $n!>n^2$ for n=4.
Let k be a positive integer [math\geq4[/tex] and suppose $k!>k^2$.
Then, $k!(k+1) > k^2(k+1)$
$(k+1)! > k^2(k+1) = k^3+k^2$

Here's where I'm stuck. I know $k\geq4$ but how can I show that $(k+1)^2 < k^3+k^2$?

From then, I can show $(k+1)! > k^3+k^2 > (k+1)^2$
Therefore, $(k+1)! > (k+1)^2$, so $n! > n^2 for n \geq4$.

Please help me out with this way of solving the proof, I'm sure there are other ways.

you will have that $k^3 + k^2 > (k + 1)^2$ if you can show $k^3 > 2k + 1$ for $k \ge 4$

3. so how do I go from $(k+1)!>k^3+k^2$ to $k^3>2k+1$ - where did the 2k+1 come from?

4. Originally Posted by cassiopeia1289
so how do I go from $(k+1)!>k^3+k^2$ to $k^3>2k+1$
well, if i told you that, i'd be doing the problem for you, wouldn't i?

- where did the 2k+1 come from?
we want to show that

$k^3 + k^2 > (k + 1)^2 = k^2 + {\color{red}2k + 1}$

thus we need to show that $k^3 > 2k + 1$