Question

\[ \frac{n!}{2^n}\]

Answer 1

\[lim_{n \rightarrow \infty}\]

Answer 2

I can see it's divergent...just want to know how to show it mathematically i guess

Answer 3

how is n!=(n+1)n?

Answer 4

it's a guess

Answer 5

a wrong guess

Answer 6

4! = 4*3*2*1= 24
(4+1)4 = 20

Answer 7

what is the question?

Answer 8

determine whether the following converges or diverges
\[a_{n} = \frac{n!}{2^n}\]

Answer 9

the limit is not 0 n! grows much faster than 2^n

Answer 10

I'm just trying to understand it
\[\frac 1 2 \frac 2 2 \frac 3 2 ...\frac{n-1}{2} \frac n 2 >= \frac 1 2 \frac n 2\]

Answer 11

I think what you could do, is use a variation of the ratio test. If you look at the limit \[\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\]

Answer 12

You get \[\lim_{n \to \infty} \frac{(n+1)!}{2^{n+1}}\frac{2^n}{n!}=\lim_{n\to\infty}\frac{n}{2}=\infty\]Therefore, it must diverge.

Answer 13

the ratio test makes perfect sense....but the solution that I wrote above from the manual doesn't make much sense to me

Answer 14

What did the manual say?

Answer 15

\[\frac 1 2 \frac 2 2 \frac 3 2 ...\frac{n-1}{2} \frac n 2 >= \frac 1 2 \frac n 2\]

Answer 16

then n/4 -> infty

Answer 17

I don't really understand what they're trying to do either...

Answer 18

are they subbing (n-1)n for n!?

Answer 19

squeeze theorem?

Answer 20

its like the squeeze theorem since something_1 greater than something_2 that is not zero, something_1 is greater than 0

Answer 21

I don't think it's the squeeze theorem at all, since they only have the lower bound. However, it's similar in that they're doing a comparison. Since \[\frac{n!}{2^n}\ge\frac{n}4\]and \(\frac{n}{4}\) diverges, so must \(\frac{n!}{4}\).

Answer 22

comparison test?

Answer 23

I believe that's what they're using.

Answer 24

I see...I prefer the ratio test though...thanks again Sir George!

Answer 25

I also prefer the ratio test. And you're welcome.