# Series Calc II

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.

## Answers and Replies

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

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

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

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

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