# Ratio Test

1. Jan 30, 2012

### fauboca

$$\displaystyle\sum_{n = 1}^{\infty}\frac{(n!)^3}{(3n)!}z^n , \ z\in\mathbb{C}$$

By the ratio test,
$$\displaystyle L = \lim_{n\to\infty}\left|\frac{[(n + 1)!]^3 z^{n + 1} (3n)!}{[3(n + 1)]! (n!)^3 z^n}\right| = \lim_{n\to\infty}\left|\frac{z (n + 1)^2}{3}\right| = \infty.$$

Therefore, the series diverges and there is no radius of convergence.

Correct?

Last edited: Jan 30, 2012
2. Jan 30, 2012

### lanedance

i don't think your cancellation is quite correct 3(n+1)=3n+3

3. Jan 30, 2012

### fauboca

Thanks, I see the problem.

$$|z|<27$$

So $R = 27$ then

Last edited: Jan 30, 2012
4. Jan 30, 2012

### lanedance

can you show how you got there?

Last edited: Jan 30, 2012
5. Jan 30, 2012

### fauboca

$$\lim_{n\to\infty}\left|\frac{z (n + 1)^3}{(3n + 3)(3n + 2)(3n + 1)}\right|$$

$$\Rightarrow \lim_{n\to\infty}|z|\left(\frac{n^3}{27n^3}\right) = \frac{1}{27}|z| < 1 \Rightarrow |z| < 27$$

6. Jan 30, 2012

### lanedance

yeah looks good

you can just just take the result below if you like , rather than carrying the z through though its good to understand why
$$r = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right|$$