ModernLogic

## Main Question or Discussion Point

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.

Pyrrhus
Homework Helper
is the index of summation k or n?

LeonhardEuler
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

Last edited:
Pyrrhus
Homework Helper
I got the exact same result, Leonhard, throught D' Alambert's Criterium (ratio test), but given he said a radius of convergence of 0 for z^n, it confused me to what is exactly the index of summation k or n.

LeonhardEuler
Gold Member
Yeah, I didn't even catch that, but its probably just a mistake.

HallsofIvy