Help please! the statement S(n) be n! >/=3^n. S(0) is true but S(1) is false.?

Question

Find the smallest number k greater than 1 such that s(k) is true. Prove by induction S(n) is true for all n>/= k

(I found out that k should be equal or greater than 7 in order for the statement to be true, but i'm having trouble of writing the proof)

Answer 1

1! = 1 which is < 3^1 = 3
2! = 2*1 = 2, which is < 3^2 = 9
3! = 3*2*1 = 6, which is less than 3^3 = 81
4! = 24, < 3^4 = 81
5! = 120, < 3^5 = 243
6! = 720, < 3^6 = 729, although this gets real close
7! = 5040, which is greater than 3^7 = 2187

So the smallest n for which this works is 7.
Now , for the induction proof:
We have established that for n = 7, that S(n)| n! >= 3^n is true.
We need to prove that for all n >= 7, that if S(n) is true, then S(n+1) is also true.

(n+1)! = (n + 1)n!
>= (n+1)(3^n), since n! >= 3^n
>= 3(3^n), since (n+1)>3 for all n>=7
>= 3^(n+1)

Therefore, for all n >= 7, S(n) | n! >= 3^(n+1) is true.

Answer 2

To prove that n! >= 3^n for n >= 7:

First, prove the base case, that it holds for n=7.
7! = 5,040, and 3^7 = 2,187, so it's true in this case.

Next, the induction step: prove that if it holds for some natural number n >= 7, then it holds for n+1.

(n+1)! = (n+1)n! >= (n+1)3^n >= 3^(n+1), because n+1 >= 8 > 3.

By the principle of induction, the statement is true for all natural numbers n>=7.

Answer 3

Assume S(k) is true for a certain natural number k > 1, that is,
k! >= 3^k

Now
(k+1)! = (k+1) k! >= (k+1) 3^k >= 3 * 3^k = 3^(k+1)

(The last step is true because k+1 >= 3.)

It follows that S(k+1) is true, which finishes induction.