- #1
fauboca
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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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