Question

prove n! >= 5^n for all n >= 12

Answer 1

so you proved it for the base case?

Answer 2

that is you proved it for n=12?

Answer 3

anyways after that
suppose k!>=5^k for some integer k>=12
and show (k+1)!>=5^(k+1) also holds.

Answer 4

My bad, yeah 12 is the base case

Answer 5

so you have no problem with that part right?

Answer 6

I get the base case yeah, and I understand I need to show (k+1)!>=5^(k+1) but I have no idea from there

Answer 7

Ok so our k's our only from 12 to bigger
so for any value k we have that k>5 right?

Answer 8

Take the inequality k!>=5^k
and see what happens if you multiply both sides by 5

Answer 9

Yes but then once you get to that part (that is 5(n!) >= 5(5^n) don't you then have to show that you're after (n+1)! >= 5(5^n)

Answer 10

5(5^n) is 5^(n+1)

now we just said k>5
if k>5 then k+1>5

Answer 11

so that means (k+1)*k!>=5*k! correct?

Answer 12

Yeah okay, and why is k>5 for any value k?

Answer 13

recall that k is bigger than or equal to 12

Answer 14

Oh I see where you're coming from

Answer 15

so k will always be more than 5

Answer 16

yep

Answer 17

since k>=12

Answer 18

So does (k+1)*k!>=5*k! prove it?

Answer 19

and write what we had on the other side of the 5*k!

Answer 20

\[(k+1)\cdot k! \ge 5 k! \ge 5(5^k)\]

Answer 21

I think I see now, you've been a great help! Thanks!

Answer 22

np