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$$\sum_{n\ge 1}\frac{(n!)^3}{(3n)!}z^{3n}=\sum_{k=3n, n\ge 1}\frac{((k/3)!)^3}{k!}z^k.$$

The ratio test seems like a better option than the root test here, but $\frac{a_{k+1}}{a_{k}}$ is either $0$ or undefined. So the radius (the limsup) is taken to be $\infty$. Am I using the ratio test correctly or should I be using it with $\frac{a_{k+3}}{a_{k}}$ instead?

José Carlos Santos
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Iced Palmer
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    The ratio tests says that the series $\sum a_n $ converges when $\lim a_{n+1}/a_n < 1$. In this case your $a_n $ includes a $z^{3n} $ term. Then solve the inequality for z. – Grant B. May 24 '18 at 06:31
  • Shouldn't $a_n$ correspond to $z^n$ instead of $z^{3n}$? – Iced Palmer May 24 '18 at 06:37
  • No, in this case it's the $n$th summand of the series. I believe you're more used to a less general form of the ratio test which is specific to power series, but the standard result examines consecutive (nonzero) terms in the series. In this series, the $n$th nonzero term contains a $z^{3n} $ – Grant B. May 24 '18 at 06:45

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Note that, if $z\neq0$,\begin{align}\lim_{n\to\infty}\frac{\left|\frac{((n+1)!)^3}{(3(n+1))!}z^{3(n+1)}\right|}{\left|\frac{(n!)^3}{(3n)!}z^{3n}\right|}&=\lim_{n\to\infty}\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}|z|^3\\&=\frac{|z|^3}{3^3}\\&=\left(\frac{|z|}3\right)^3.\end{align}Therefore, the radius of convergence is $3$.

José Carlos Santos
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