Show that $n! ≥ 2^{n-1}$ is true by induction for n being a positive integer

Question

I start off by doing the base case:

Base Case: Let P(n) be $n! \ge 2^{n-1}$. Then $P(1)$ will be:

$$ 1! \ge 2^0 $$ $$ 1 \ge 1 $$ So the base case is true.

Induction: I assume $P(k)$ is true for some arbitrary positive integer $k$. Now I need to show that $P(k+1)$ is true. So assuming $k! \ge 2^{k-1} $ is true, I need to show that $(k+1)! \ge 2^{(k+1)-1} $ is true.

$$ 2^{(k+1)-1} = 2 \cdot 2^{k-1} $$

$$\le 2 \cdot k! \tag{By inductive hypothesis}$$

My question So I have solved this to this point but I don't really know where to take it from here. I would appreciate any help on completing this proof.

Does this answer your question? Prove the inequality $n! \geq 2^n$ by induction. This other question asks for a somewhat stronger condition, so it starts at $n = 4$ instead of $n = 1$, but otherwise the induction procedure to follow is basically the same. — John Omielan, Jan 15 '21 at 07:09

score 2 · Answer 1 · answered Jan 15 '21 at 07:03

2

From $k! \ge 2^{k-1}$ and $k+1 \ge 1$ we get

$$(k+1)!=k!(k+1) \ge 2^{k-1}(k+1) \ge 2^{k-1} \cdot 2= 2^k.$$

answered Jan 15 '21 at 07:03

Fred

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Show that $n! ≥ 2^{n-1}$ is true by induction for n being a positive integer

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