Question

I am trying to get the Limit by squeeze theorem of :

(2n)!/ (n^2)^n as n goes to infinity

Answer 1

so right now I have \\[0 \le \frac{(2n)!}{(n^2)^n}\] for all n but I can't seem to find and upper bound . I tried using the ratio test but that didn't work I think squeeze theorem is the best but I can't find an upper one to compare .

Answer 2

I can't tell what the symbols are in the comment section.

Answer 3

right now I have the lower bound so I know 0 <= (2n)!/ (n^2)^n and I need an upper bound that converges to 0 to use the squeeze theorem

Answer 4

Try working with these two>
n^-2n(2n)!=>0, that's providing that n>0
Then I have,
0<=(n^4)^-n/2(2n)!
That's only if you assume n is real.

Answer 5

Do you already know it converges even?

Answer 6

According to Wolfram alpha the limit is 0 as n goes to infinity

Answer 7

and this is a sequence?

Answer 8

*series

Answer 9

this is a sequence .

Answer 10

I am trying to see if \[ ((n-1)^2)^n \le (n^2)^n \] and then flipping would give me something

Answer 11

Yeah, this whole time I was treating it as aseries, my bad.

Answer 12

\[\frac{(2n)!}{(n^2)^n}=\frac{2n(2n-1)(2n-2)\cdots 3\cdot 2\cdot 1}{n\cdot n\cdot n\cdots n\cdot n\cdot n}=\frac{(2n\cdot 1)}{n^2}\frac{((2n-1)\cdot 2)}{n^2}\cdots\frac{(n+1)n}{n^2}\]

Answer 13

Use the following theorem: If \(\sum a_n\) convergent, then \(\lim_{n\to \infty} a_n=0\)

Answer 14

@Zarkon how do you get \[ \frac{((2n-1)2)}{n^2}...\frac{(n+1)}{n^2}\] I get that you are expanding the factorial as welll as the denominator n times n^2 but the numerator? It works great after though

Answer 15

ah so you are pairing then so then you will end up with n terms and can divide exactly by n^2 n times?

Answer 16

yes (with a limit that clearly goes to zero)