Hi folks. I need to find the radius of convergence of this series: $$\sum_{k=0}^\infty \frac{(n!)^3z^{3n}}{(3n)!}$$

The thing throwing me off is the $$z^{3n}$$. If the series was $$\sum_{k=0}^\infty \frac{(n!)^3z^n}{(3n)!}$$ I can show it has radius of convergence of zero. But $$z^{3n}$$ means its only taking power multiples of 3. Does that change anything?

Thanks.

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 Recognitions: Homework Help is the index of summation k or n?
 Recognitions: Gold Member The problem can still be solved using the ratio test: $$\lim_{n\rightarrow\infty}\frac{((n+1)!)^3z^{3n+3}}{(3n+3)!}\times\frac{ (n!)^3}{z^{3n}(3n)!}$$ =$$\lim_{n\rightarrow\infty}\frac{(n+1)^3z^{3}}{(3n+3)(3n+2)(3n+1)}$$ Now, being sloppy so that I don't have to write so much, in the limit this is going to be equal to: $$\lim_{n\rightarrow\infty}\frac{(n)^3z^{3}}{27(n)^3}$$ =$$\lim_{n\rightarrow\infty}\frac{z^{3}}{27}$$ =$$\frac{z^{3}}{27}$$ so, requiring the absolute value of this expression to be less than 1 gives a radius of 3

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