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I need to find the p value for which every n: $n \ge p$ and $n \in \mathbb{N}$

I am given:

$n! \ge 2^n$

I'm really not sure how to approach this question. Say my base proof is $p=4$ so $n \ge 4$ and then I tried:

$(n+1)! \ge 2^n \times 2^1$ (what was originally $2^{n+1}$)

$n! \times (n+1) \gt 2^n \times 2^1$

But this is where I get stuck.. By the way, is there any other way to find $p$ other than just trying numbers?

Thanks!

ericw31415
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bm1125
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  • Where does $n!\ge 2^n$ come from? – Shuri2060 Aug 07 '17 at 16:29
  • @Shuri2060 I believe it's the problem statement ("Find $p$ for which, $\forall n \geq p$, $n! \geq 2^n$"). – platty Aug 07 '17 at 16:31
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    If $n!\geq2^n$ and $n\geq4$ then $(n+1)!=n!\times(n+1)\geq2^n\times4>2^n\times2^1=2^{n+1}$. – drhab Aug 07 '17 at 16:33
  • See answers to this question about how to write a clear induction proof. Do not fall into the (very bad) habit of starting with what you hope is true and then making simplifications to reach a tautology. Instead, start with what you know is true and use that to prove what you hope is true. – JMoravitz Aug 07 '17 at 16:49
  • @platty is correct. Thanks everyone . I will try to write clear induction next time! – bm1125 Aug 08 '17 at 07:22

2 Answers2

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If you take $p=4$, then you have $p!=24>16=2^p$, so you only need to prove that for $n\geq 4$, you have $n!\geq 2^n$.

So, you already have step 1 of the induction proof, so let's suppose that you have for $k\geq 4$ that

$$k!\geq2^k.$$

Then

$$(k+1)!=k!(k+1)\geq2^k(k+1)\geq2^k(4+1)\geq2^k2=2^{k+1}.$$

Sergio Enrique Yarza Acuña
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$$p=1, 1!=1 <2^1$$

$$p=2, 2!=2 <2^2$$

$$p=3, 3!=6 <2^3$$

$$p=4, 4!=24>2^4$$

So, for $n\ge 4$ we can prove using induction that $n!\ge 2^n $.

$$(n+1)!=n! (n+1)\ge 2^n (n+1) $$ but $$n\ge 4\implies (n+1)\ge 2$$ thus $$(n+1)!\ge 2^n.2=2^{n+1} $$

Done!.

hamam_Abdallah
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