- #1

- 20

- 0

I know you have to do the ratio test for this one but I don't know how to do it.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Calculus!
- Start date

- #1

- 20

- 0

I know you have to do the ratio test for this one but I don't know how to do it.

- #2

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,566

- 23

[tex]s_n=\frac{n!^{{2a}^n}}{(2n)!}[/tex]

Can you write down [itex]s_{n+1}[/itex]?

- #3

- 20

- 0

im sry its (n!)^2(a^n) for the numerator

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

[tex]\left[\frac{((n+1)!)^2a^{n+1}}{(2(n+1))!}\right]\left[\frac{(2n)!}{((n!)^2a^n}\right]= \left[\frac{(n+1)!}{n!}\right]^2\left[\frac{a^{n+1}}{a^n}\right]\left[\frac{(2n)!}{(2(n+1))!}][/tex]

[tex]= \frac{(n+1)(a)}{(2n+2)(2n+1)[/tex]

What is the limit of that as n goes to infinity? If that limit depends on a, then the series will converge only for values of a that make the limit less than 1.

Share: