elem.num.theo - help w/proof

**cassiopeia1289** · September 25th 2008, 01:43 PM

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.

any help: please!!!??

**Jhevon** · September 25th 2008, 08:22 PM

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.

any help: please!!!??