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

By the ratio test,

[tex]

\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.[/tex]

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

Correct?

By the ratio test,

[tex]

\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.[/tex]

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

Correct?

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