# elem.num.theo - help w/proof

• Sep 25th 2008, 01:43 PM
cassiopeia1289
elem.num.theo - help w/proof
So - I know all the steps, I just need help with a gap.

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

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

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

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

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

• Sep 25th 2008, 08:22 PM
Jhevon
Quote:

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

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

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

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

From then, I can show \$\displaystyle (k+1)! > k^3+k^2 > (k+1)^2\$
Therefore, \$\displaystyle (k+1)! > (k+1)^2\$, so \$\displaystyle 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 \$\displaystyle k^3 + k^2 > (k + 1)^2\$ if you can show \$\displaystyle k^3 > 2k + 1\$ for \$\displaystyle k \ge 4\$
• Sep 26th 2008, 05:49 AM
cassiopeia1289
so how do I go from \$\displaystyle (k+1)!>k^3+k^2\$ to \$\displaystyle k^3>2k+1\$ - where did the 2k+1 come from?
• Sep 26th 2008, 06:18 AM
Jhevon
Quote:

Originally Posted by cassiopeia1289
so how do I go from \$\displaystyle (k+1)!>k^3+k^2\$ to \$\displaystyle k^3>2k+1\$

well, if i told you that, i'd be doing the problem for you, wouldn't i? :p

Quote:

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

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

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