Finding Radius of Interval Convergence: \sum x^n/2^n

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SUMMARY

The discussion focuses on finding the radius of interval convergence for the series \(\sum_{n=1}^{\infty} \frac{x^n}{2^n}\). Participants recommend using the ratio test as a primary method, stating that it is commonly employed for power series. However, they also suggest that the root test may provide a simpler solution for this specific series. Both tests involve limits to determine convergence, with specific conditions outlined for absolute convergence and divergence.

PREREQUISITES
  • Understanding of power series and their convergence properties
  • Familiarity with the ratio test for series convergence
  • Knowledge of the root test for series convergence
  • Basic calculus concepts, including limits
NEXT STEPS
  • Study the application of the ratio test in detail, particularly for power series
  • Explore the root test and its advantages in specific scenarios
  • Practice finding the radius of convergence for various power series
  • Investigate other convergence tests, such as the comparison test and the integral test
USEFUL FOR

Students studying calculus, particularly those focusing on series and convergence, as well as educators teaching these concepts in mathematics courses.

Lance WIlliam
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Homework Statement



Finding the Radius of interval convergence of \sum n=1(there's a infinity on the sigma), "x^n/2^n"

I really don't have a clue on which way I should go.
Just a hint would be great:)

Homework Equations





The Attempt at a Solution

 
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Hint: ratio test.
 
You almost always use the ratio test to find the radius of convergence of a power series. For this particular problem you may find that the root test is simpler. But for most power series the ratio test is simplest.

Ratio test: The series \sum a_n converges absolutely if
\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}&lt; 1[/itex]<br /> It diverges if that limit is larger than one and may converge absolutely, converge conditionally, or diverge if the limit is equal to 1.<br /> <br /> Root test: The series \sum a_n converges absolutely if <br /> \lim_{n\rightarrow \infty}\left( ^n\sqrt{a_n}\right)&amp;lt; 1[/itex]&lt;br /&gt; It diverges if that limit is larger than one and may converge absolutely, converge conditionally, or diverge if the limit is equal to 1.
 
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