What is the convergence of a power series using the ratio test?

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SUMMARY

The discussion focuses on applying the ratio test to determine the convergence of a power series. The user presents a specific problem involving the series with terms including \((-1)^n\) and \(|x+2|/2\). The conclusion reached is that the radius of convergence is 2, specifically for the interval \((4, 0]\). The user successfully simplifies the expression by recognizing that the alternating factor \((-1)^n\) does not affect the convergence analysis.

PREREQUISITES
  • Understanding of power series and their convergence criteria
  • Familiarity with the ratio test for series convergence
  • Knowledge of limits and the behavior of sequences
  • Basic algebraic manipulation of series terms
NEXT STEPS
  • Study the detailed application of the ratio test in various contexts
  • Learn about the properties of alternating series and their convergence
  • Explore the concept of radius and interval of convergence for power series
  • Investigate the implications of the nth term test in series analysis
USEFUL FOR

Students studying calculus, particularly those focusing on series convergence, as well as educators looking for examples of applying the ratio test in power series problems.

rcmango
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Homework Statement



I've tried to apply the ratio test to a problem that is a power series. here's the problem as a pic: http://img152.imageshack.us/img152/2751/35685690oj3.png

Homework Equations





The Attempt at a Solution



I've gotten so far as you can see in the pic, I've skipped all my work, but if someone can go through the problem, you'll see where I'm stuck, I'm not sure what to do with the n/(n+1) ...just use nth term test.. to get 1?

then also, |x+2|/2 ... I'm unsure if it just stays that way?

and the big problem is my (-1)^n i can't divide that through the numerator because of the n right?

there is my effort, now please just help me to achieve the end, i don't have much time.

i know the answer is 2, (4, 0]
 
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\frac{n}{n+1}*\frac{2^n}{2^{n+1}}*\frac{(x+2)^{n+1}}{(x+2)^n}

I get this if I just skip the (-1)^n part cos you just know it is 1, -1, 1, -1 for different values of n which tells you that the sum is alternating.
 

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