You can easily show by induction that for n>6, n! > 3^n. Then .
Post again in this thread if you are still having trouble.
Is there a way to show my work for this? (other than knowing that both will grow infinitely and by pluggin in numbers to see which one grows faster): http://www.wolframalpha.com/input/?i=lim+as+n-%3Einf+((n!)%2F(2^n))
Any input would be greatly appreciated!
Thanks in advance!
I'll start by saying I like Soroban's answer. It's more clear and direct. However, if you don't know about the ratio test...
By induction, I mean the method of proof where you prove some statement P(n) for n=1, then prove that P(n+1) follows from P(n). You can then conclude that P(n) is true for all natural numbers.
So, if n=7, n!=5040 and 3^n=2187, so in this case n! > 3^n.
Assuming that for some n, n! > 3^n, since n > 6 > 2, n+1 > 3 and we can multiply the inequalities, giving (n+1)! > 3^(n+1). (Note that the ratio test is hidden in the previous line of reasoning).
So n! > 3^n for all n > 6. Of course, I chose n > 6 because 7 is the first integer for which it is true: 6! = 720 is not greater that 3^6 = 729.
I hope that helps. Go ahead and post again in this thread if you're still having trouble.
- Hollywood