3
$\begingroup$

Prove $n^2 < n!$.

This is what I have gotten so far

basis step: $p(4)$ is true Inductive Hypothesis assume $p(k)$ true for $k \ge 4$

Inductive Step $p(k+1)$ : $(k+1)^2 < (k+1)!$

$$(k+1)^2 =k^2 + 2k + 1 < k! + 2k +1$$

Can someone please explain the last step this is from text, I need help understanding this, forgive me for the formatting error Im still learning

$\endgroup$
5

2 Answers 2

4
$\begingroup$

Inductive Step:

Assume the case for $n$ is true, then for $n \geq 4$ $$(n + 1)^2 = n^2 + 2n + 1 < n! + 2n + 1 < n! + n^2 \leq n! + n!n = n!(n+1) = (n+1)!.$$

$\endgroup$
1
  • 1
    $\begingroup$ @MauroALLEGRANZA, I misread! I fix rright now. This is what I get for being lazy. $\endgroup$
    – IAmNoOne
    Sep 20, 2014 at 18:20
1
$\begingroup$

$$n^n \geq n!$$

Proof: Let $n\in\mathbb{N}$. Then $$ n^n = n\cdot n\cdot n\cdot...\cdot n$$ where as $$ n! = n\cdot(n-1)\cdot (n-2)\cdot...\cdot1$$

For each term in the product you can compare $$n = n $$ $$n > n-1 $$ $$n > n-2 $$ and so on. Thus $n^n \geq n!$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.