Limsupremum calculation horrible difficulty

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SUMMARY

The discussion centers on calculating the radius of convergence for the series \(\sum_{n=0}^\infty a_{n}z^{2n}\) given that \(\sum_{n=0}^\infty a_{n}z^{n}\) has a radius \(R\). Participants emphasize using the formula \(\limsup|a_n|^{\frac{1}{n}}=\limsup |\frac{a_{n+1}}{a_n}|\) to derive the radius of convergence. It is established that the transformation of \(z\) to \(z^2\) affects the convergence behavior, specifically indicating that the new radius of convergence is \(\sqrt{R}\).

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the concept of radius of convergence
  • Knowledge of the \(\limsup\) and its application in series
  • Basic proficiency in LaTeX for mathematical expressions
NEXT STEPS
  • Study the derivation of the radius of convergence using the ratio test
  • Explore the implications of transforming variables in power series
  • Learn about the properties of \(\limsup\) in the context of sequences
  • Practice writing mathematical expressions in LaTeX for clarity
USEFUL FOR

Students in advanced calculus, mathematicians working with power series, and educators teaching convergence concepts will benefit from this discussion.

betty2301
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Homework Statement


Calculate the radius of convergence of [itex]\sum_{n=0}^\infty a_{n}z^{2n}[/itex]
let [itex]\sum_{n=0}^\infty a_{n}z^{n}[/itex]with radius R

Homework Equations


[itex]\limsup|a_n|^{\frac{1}{n}}=\limsup |\frac{a_{n+1}}{a_n}|[/itex]

The Attempt at a Solution


[tex]\limsup|a_n|^{\frac{1}{n}}=\lim_{k\to\infty}|a_{2k}|^{\frac{1}{2k}}[/tex]
then how to write the "ratio"....
the latex is killin me please help....
 
Last edited:
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You could use the limsup formula, but it's easier to think about what the radius of convergence means. Basically you're given that the original series converges for |z| < R and diverges for |z| > R, so what happens when you square z?
 

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