Inequality induction proof

Question

I have been practicing using Mathematical Induction, in proofs. I came across a problem in my practice problems list that is giving me a lot of trouble. This is the question

Prove that $n! > n^3\ \mbox{for}\ n > 5$

So here is my inductive step that I have done so far.

assumption, $k! > k^3$ for some $n = k$

$$n = k+1$$

$$ (k+1)! = (k+1)k! > (k+1)k^3 $$

After this I know that I have to show somehow that $(k+1)k^3 > (k+1)^3$, but I have no ideas on how to do this. Can anyone give me a hint on what to do next? Thanks for the help!

score 0 · Accepted Answer · answered Sep 22 '14 at 01:25

0

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

Notice for $k > 5 $,

$$ 3k^2 + 3k + 1 < 3k^2 + 3k^2 + k^2 = 7k^2 < k^4. $$

answered Sep 22 '14 at 01:25

Why only k^4 why not k^4 + k^3, the rest I understand perfectly. – ultrainstinct Sep 22 '14 at 01:31
Notice that my expression $3k^2 + 3k + 1$ is missing the $k^3$ from the line above. Thus we have $$ (k + 1)^3 = 3k^2 + 3k + 1 + k^3 < k^4 + k^3.$$ – Sep 22 '14 at 01:31
Yes I see this now, wow thanks for the help! – ultrainstinct Sep 22 '14 at 01:38
One question though, I can't just state that, I need to prove that too by induction right? – ultrainstinct Sep 22 '14 at 12:08
If $ k > 5 $ then $ k^2 > 25 > 7 $. – Sep 22 '14 at 15:48

Inequality induction proof

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