Does the exponent in a series affect its radius of convergence?

[ModernLogic]

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]

is the index of summation k or n?

 

[LeonhardEuler]

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

 

[Pyrrhus]

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]

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

 

[HallsofIvy]

Modern Logic: The only difference between having z³ⁿ instead of zⁿ in the problem is that you will have z³ instead of z in the final formula- Take the cube root. (Let y= z³ so that the sum involves yⁿ.)

However, you are wrong when you say that if the problem involved zⁿ instead of z³ⁿ you would get a radius of convergence of 0. As Leonard Euler said, you would get 27 and so for z, the cube root of that, 3.