Hi, can someone help me prove this inequality by Induction
The inequality is: n! ≥ 2^{n-1 }for all natural numbers n
Thank You
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 that if Statement(k) is true, then Statement(k+1) is true.
Therefore, you've proven Statement(n) is true for all n>=1.
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 $\displaystyle k \ge 1, \text{ you know that } k+1 \ge 2.$