Determining the raidus of convergence

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SUMMARY

The radius of convergence for the power series Ʃ(x^(2n))/n! is determined using the ratio test, yielding a limit of 0 as n approaches infinity. This indicates that the series converges for all values of x, resulting in an infinite radius of convergence. The relationship between the ratio r and the radius R is established as R = 1/r, confirming that as r approaches 0, R approaches infinity.

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  • Understanding of power series and their convergence properties
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  • Basic knowledge of factorial notation and limits
  • Concept of radius of convergence in mathematical analysis
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Homework Statement


Determine the radius of convergence of the given power serie .

Ʃ(x^(2n))/n!

n goes from 0 to infinity

Homework Equations


limit test ratio


The Attempt at a Solution


I am using the limit test ratio and I've got this : [n! * x^(2n+2)]/[(n+1)! * x^2n], then [n!* x^2n * x^2]/[(n+1) * n! * x^2n] , canceling the common things I am left with lim n-> infinity of x^2/n+1, which is 0, but the radius of convergence is infinity, why is infinity?
 
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bigu01 said:

Homework Statement


Determine the radius of convergence of the given power serie .

Ʃ(x^(2n))/n!

n goes from 0 to infinity

Homework Equations


limit test ratio


The Attempt at a Solution


I am using the limit test ratio and I've got this : [n! * x^(2n+2)]/[(n+1)! * x^2n], then [n!* x^2n * x^2]/[(n+1) * n! * x^2n] , canceling the common things I am left with lim n-> infinity of x^2/n+1, which is 0, but the radius of convergence is infinity, why is infinity?

Because if the ratio test gives you a ratio of r then the corresponding radius of convergence is R=1/r. So as r gets very small the radius of convergence gets very large. If you go all the way to r=0 then R=infinity.
 
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