Radius of Convergence

  • Thread starter ModernLogic
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  • #1
ModernLogic

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

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

Thanks.
 

Answers and Replies

  • #2
Pyrrhus
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is the index of summation k or n?
 
  • #3
LeonhardEuler
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The problem can still be solved using the ratio test:
[tex]\lim_{n\rightarrow\infty}\frac{((n+1)!)^3z^{3n+3}}{(3n+3)!}\times\frac{(n!)^3}{z^{3n}(3n)!}[/tex]
=[tex]\lim_{n\rightarrow\infty}\frac{(n+1)^3z^{3}}{(3n+3)(3n+2)(3n+1)}[/tex]
Now, being sloppy so that I don't have to write so much, in the limit this is going to be equal to:
[tex]\lim_{n\rightarrow\infty}\frac{(n)^3z^{3}}{27(n)^3}[/tex]
=[tex]\lim_{n\rightarrow\infty}\frac{z^{3}}{27}[/tex]
=[tex]\frac{z^{3}}{27}[/tex]
so, requiring the absolute value of this expression to be less than 1 gives a radius of 3
 
Last edited:
  • #4
Pyrrhus
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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.
 
  • #5
LeonhardEuler
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Yeah, I didn't even catch that, but its probably just a mistake.
 
  • #6
HallsofIvy
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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.
 

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