# Series Calc II

1. May 1, 2009

### 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.

2. May 1, 2009

### Tom Mattson

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

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

Can you write down $s_{n+1}$?

3. May 1, 2009

### Calculus!

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

4. May 1, 2009

### HallsofIvy

Staff Emeritus
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.