What is the Radius of Convergence for ∑[(2n+1)/2n] x^n?

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SUMMARY

The radius of convergence for the power series ∑_(n=0)^∞[(2n+1)/2n] x^n is determined using the Ratio Test. By applying the Ratio Test, specifically the limit lim_n->inf (a_n+1 / a_n), one can simplify the expression to find the radius. The discussion emphasizes the importance of obtaining a second-degree polynomial in both the numerator and denominator for effective analysis. Additionally, the application of L'Hôpital's rule is suggested to resolve complex fractions encountered during the calculation.

PREREQUISITES
  • Understanding of power series and convergence
  • Familiarity with the Ratio Test for series convergence
  • Knowledge of L'Hôpital's rule for limit evaluation
  • Basic algebraic manipulation of polynomials
NEXT STEPS
  • Practice applying the Ratio Test on various power series
  • Study the application of L'Hôpital's rule in limit problems
  • Explore the concept of radius of convergence in more complex series
  • Review polynomial long division for simplifying fractions
USEFUL FOR

Students studying calculus, particularly those focusing on series and convergence, as well as educators seeking to enhance their teaching of power series concepts.

kgarcia3
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Homework Statement



Find the radius of convergence of the following power series: ∑_(n=0)^∞[(2n+1)/2n] x^n)

Homework Equations



Ratio Test: lim_n->inf (a_n+1 / a_n)

The Attempt at a Solution



I got a big ugly fraction that involved both n and x
 
Last edited:
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Show us what you got.
 
kgarcia3 said:

Homework Statement



Find the radius of convergence of the following power series: ∑_(n=0)^∞[(2n+1)/2n] x^n)

Homework Equations



Ratio Test: lim_n->inf (a_n+1 / a_n)

The Attempt at a Solution



I got a big ugly fraction that involved both n and x

You should be able to get a second degree polynomial on the numerator and denominator... then apply L'Hopital rule.. why not show us the fraction you've gotten
 

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