To find the formula you can compute the first terms of the series.
Another way is to use
Could you verify that I am on the right track for another?
Prove that 3^n < n! if n is an integer greater than 6..
- Base case 3^7 = 2187 and 7! = 5040
- Assume for p(k) show for p(k+1)
- 3^K+1 < K+1!
- 3^k + 3^k < K + 1!
I am stuck on this part now.
You should have an initial condition that n>2
Let the following is true
Putting k! instead of 3^k we get something greater hence if we can prove that this greater thing less than our RHS
then we have done it