Finding the Largest Value of b for Convergence

[Calculus!]

Find the largest value of b that makes the following statement true: "if 0<= a <= b, then the series (from n=1 to infinity) of (((n!)^2a^n)/(2n!)) converges".

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

 

[quantumdude]

OK let's call the summand s_n. So we have:

s_n=\frac{n!^{{2a}^n}}{(2n)!}

Can you write down s_{n+1}?

 

[Calculus!]

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

 

[HallsofIvy]

Yes, use the ratio test.
\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))!}]
= \frac{(n+1)(a)}{(2n+2)(2n+1)
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.