Question

Prove that n! > n^2 for every integer n>= 4 and n! > n^3 for every integer n>= 6.
Question is from induction chapter.

Answer 1

In the example about Luca's sequence proof they used induction after dividing the number into the ones that it is made up of, however, I can't think of a way how to apply this technique to factorials...

Answer 2

if n = k
assume
k! >= k^2 - which it is for k = 4
Now we need to show that when n = k+1:-
(k+1)k! >= (k+1)^2

Answer 3

Now can we derive that last one ?

Answer 4

sorry that should be > not >=

Answer 5

Hm... But how would it be possible to derive it? I understand the very basics, but I can't seem to understand how to get from k! > k^2 to (k+1)! > (k+1)^2 normally I would add something to both sides, but it's not a sequence, so I am having trouble...

Answer 6

Oh, I see that it's possible to multiply both sides by (k+1), so the left side would become (k+1)! and right one - k^2(k+1), as you had said @welshfella. However, what would be the next step?

Answer 7

what if we multiplied the first equation by (k + 1)

Answer 8

lol1 - we both thought of that the same time!

Answer 9

well since k is a positive number
k^3 + k^2 is greater than (k + 1)^2

Answer 10

Base Case = n!>n² => H(k)
Assuming H(k) is true
n!>n²
n! x n+1>n² x n+1
(n+1)!>n³+n²>(n+1)² for n>=4
H(k) => H(k+1)
H(k) is true (prove that by plugging 4 into the base case)

Answer 11

And exactly the same strategy for the second part.

Answer 12

Yes - thats basically the same reasoning.

Answer 13

Oh! So, by proving that it is true for bigger number we proved that it is true for the lesser one... Interesting! Thank you all for help! :)
(Not sure to whom should I give the medal now...)

Answer 14

Welshfella is the worthy candidate

Answer 15

its up to you. I dont worry much about medals

Answer 16

Okay, then Welshfella. Thank you again!

Answer 17

yw