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I have the following two relations:

2^x =< (x+2)!

and

2^(x+1) =< (x+3)!

I assumed that the first statement is true and now, I'm trying to prove that the second statement is true too. I understand that I will have to substitute the first relation in the second and I attemted this multiple times, but I don't reach a relation that will prove it. Thanks

Jason Macville
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  • If you have proven $2^x \le (x+2)!$ then replace $x$ with $[x+1]$ and the second result immediately follows. $2^{[x+1]} \le ([x+1] + 2)!$.. – fleablood Oct 19 '17 at 21:25
  • Oh, you assumed it was true and you are trying to do a proof by induction.... I see. Okay. $2^{x+1} = 22^x \le 2(x+2)! < 3*(x+2)! \le (x+3)(x+2)! = (x+3)!$. – fleablood Oct 19 '17 at 21:28

2 Answers2

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Most likely this is a proof involving induction over the natural numbers, so I will simply walk through the proof.

$\underline{\text{Proposition}}:$ For all $n \in \mathbb{N}$ it is the case that $2^n \leq (n+2)!$

$\underline{\text{Proof}}:$ For our base case, notice that if $n = 0$ that $2^0 = 1 < 2 = 2! = (0 + 2)!$.

Now for our inductive step assume for some $k \in \mathbb{N}$ that $2^k = (k+2)!$ (our inductive hypothesis), and we may then see that \begin{align*} 2^{k+1} &= 2 (2^k) \\ &\leq (k+3) 2^k \text{ (as }k + 2 \geq 2 \text{ for all natural numbers } k \text{)}\\ &\leq (k+3)((k+2)!) \text{ (by our inductive hypothesis)} \\ &= (k+3)! \\ \end{align*} Thus we know that $2^{k+1} \leq ((k+1)+2)!$ as desired.

Therefore, by mathematical induction we have shown that for all $n \in \mathbb{N}$ that $2^n \leq (n+2)!$

$$\tag*{$\blacksquare$}$$

Aidan W. Murphy
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Assuming $x$ is a non-negative integer and $2^x \leq (x + 2)!$

\begin{align*} (x + 3)! &= (x + 3) (x + 2)! \\ &\geq 2 (x + 2)! \\ &\geq \ldots \end{align*}

Tsuret
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