Is there a way to show my work for this limit?

• Apr 3rd 2010, 07:16 PM
s3a
Is there a way to show my work for this limit?
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!
• Apr 3rd 2010, 07:51 PM
hollywood
You can easily show by induction that for n>6, n! > 3^n. Then $\displaystyle \lim_{n \to \infty}\frac{n!}{2^n}>\lim_{n \to \infty}\frac{3^n}{2^n}=\infty$.

Post again in this thread if you are still having trouble.
• Apr 3rd 2010, 07:54 PM
Soroban
Hello, s3a!

Quote:

Converge or diverge? .$\displaystyle \lim_{n\to\infty}\frac{n!}{2^n}$
Did you try the Ratio Test?

. . $\displaystyle R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!}{2^{n+1}}\cdot\frac{2^n}{n!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{2^n}{2^{n+1}} \;=\;\frac{n+1}{2}$

Then: .$\displaystyle \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\frac{n+1}{2} \;=\;\infty$

. . The sequence diverges.

• Apr 3rd 2010, 09:14 PM
s3a
What's the induction method?
• Apr 4th 2010, 12:13 AM
hollywood
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