Prove $n2^n < n!$ for $n \geq 6$

Question

So far I have this: Let P(n) be the statement "$n2^n \lt n!$". $k_{0}=6$. $(6)2^6 = 384 <720=6!$. $P(k_{0})$ is true. Let $n \geq 6$ and assume P(n) to be true. By the induction hypothesis, $(n+1)2^{n+1}=(n+1)(2)2^n ...$ Somehow this gets to be $n!(n+1)(2)2^n$. I am clearly missing a few steps in my proof.

Answer 1

The key step is the last step: $(n+1)! = (n+1)n! > (n+1)(n\cdot 2^n)\ge(n+1)(2\cdot 2^n) = (n+1)2^{n+1}$. I hope this helps.

Answer 2

Here's what you do. You start off by assuming that $$n2^n \lt n!$$ for some number $n_0$. Then $$n_02^{n_0} \lt n_0!$$ By multiplying both sides by $n_0+1$, we get $$n_0(n_0+1)2^{n_0} \lt (n_0+1)!$$ Now notice that $n_0(n_0+1)2^{n_0} \gt (n_0+1)2^{n_0+1}$ whenever $n_0 \gt 2$, so $$(n_0+1)2^{n_0+1} \lt (n_0+1)!$$ Which proves the statement for all $n \ge 6$, since you already established it for $n=6$.

Answer 3

Assume this holds:

$$ n2^n < n!\qquad(1) $$

Multiplying (1) by $n+1$:

$$n(n+1)2^n < (n+1)!\qquad(2)$$

But $n > 2$ so from (2):

$$(n+1)2*2^n < n(n+1)2^n < (n+1)!$$