Does the Complex Series Sum of (n!)^3/(3n)! * z^n Diverge?

[fauboca]

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

By the ratio test,
<br /> \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?

 

[lanedance]

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

 

[fauboca]

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

Thanks, I see the problem.

|z|&lt;27

So R = 27 then

 

[lanedance]

can you show how you got there?

 

[fauboca]

can you show how you got there?

\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| &lt; 1 \Rightarrow |z| &lt; 27

 

[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|