Hi, can someone help me prove this inequality by Induction

The inequality is: n! ≥ 2^{n-1 }for all natural numbers n

Thank You

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- Oct 29th 2012, 10:10 AMedexcelgceProof by Induction
Hi, can someone help me prove this inequality by Induction

The inequality is: n! ≥ 2^{n-1 }for all natural numbers n

Thank You - Oct 29th 2012, 10:38 AMjohnsomeoneRe: Proof by Induction
1) Define the statement that you'll prove by induction holds for every n:

$\displaystyle \text{Statement(n) is }"n! \ge 2^{n-1}".$

2) Show Statement(1) is true.

3) ASSUME Statement(k) is true for some k >=1. From that algebracially manipulate it so that you can show that Statment(k+1) must also be true.

It should look like this:

$\displaystyle \text{ASSUME } k! \ge 2^{k-1} \text { for some integer } k \ge 1.$

$\displaystyle \text{Then ... (insert your algebra here - look at the final result you're trying to get to)... }$

$\displaystyle \text{Therefore }(k+1)! \ge 2^{k}.$

4) 1-3 completes the proof by induction:

You've shown Statement(1) is true by #2,

and, by #3, you've shown thatStatement(k) is true,**if**Statement(k+1) is true.**then**

Therefore, you've proven Statement(n) is true for all n>=1. - Oct 30th 2012, 01:58 AMedexcelgceRe: Proof by Induction
Can you please show me the algebra that occurs at step

"Then ... (insert your algebra here - look at the final result you're trying to get to)

Thanks - Oct 30th 2012, 11:03 AMjohnsomeoneRe: Proof by Induction
I'll give you a huge hint.

You have $\displaystyle k! \ge 2^{k-1} \text{ for some } k \ge 1.$

You want to manipulate that to show that you also have that $\displaystyle (k+1)! \ge 2^{k}.$

What do you have to do to each side of the original inequality to get to this desired outcome?

You'll use that, since $\displaystyle k \ge 1, \text{ you know that } k+1 \ge 2.$