# Don't understand this For which nonnegative integers n is nē < n!? prove your answer?

• Mar 2nd 2009, 08:57 PM
Grillakis
Don't understand this For which nonnegative integers n is nē < n!? prove your answer?
This problem been giving me fits;

Q: For which nonnegative integers n is nē < n!? prove your answer?
__________________________________________________ ______________

I know I have to use Basis Step, Inductive Hypothesis and Inductive Step. But when doing the basis step, if you let n > 1, then n^(2) is larger than n!, so how can the Basis hold up? I though Basis is when the LHS is equal to the RHS or it has to be true like 4 > 2.

This is what I am stuck at?
• Mar 2nd 2009, 11:03 PM
CaptainBlack
Quote:

Originally Posted by Grillakis
This problem been giving me fits;

Q: For which nonnegative integers n is nē < n!? prove your answer?
__________________________________________________ ______________

I know I have to use Basis Step, Inductive Hypothesis and Inductive Step. But when doing the basis step, if you let n > 1, then n^(2) is larger than n!, so how can the Basis hold up? I though Basis is when the LHS is equal to the RHS or it has to be true like 4 > 2.

This is what I am stuck at?

Consider the function:

$f(k)=k^2(k+1)-(k+1)^2=k^3-2k+1$

$k \in \mathbb{R}$ , then:

$f'(k)=3k^2-2$

Now if $k\ge 1,\ f'(k)>0$ and so $f(k)$ is increasing, and as $f(1)=0,\ f(k)\ge0$ for all $k\ge 1$

Hence for all natural numbers $k\ge 1$ we have:

$f(k)=k^2(k+1)-(k+1)^2 \ge 0$

or:

$k^2(k+1)\ge (k+1)^2 \ \ \ \ \ \ \ \ \ \ \ ...(1)$

Now for the induction step of the proof assume that for some integer $k>1$ that $k^2\le k!$, then:

$k^2(k+1)\le (k+1)!$

but by $(1)$ above:

$(k+1)^2 \le k^2(k+1)\le (k+1)!$

etc...

CB
• Mar 3rd 2009, 01:56 PM
Grillakis
I see...thanks CB