Find the interval of convergence for the given power series.

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SUMMARY

The discussion focuses on finding the interval of convergence for the power series defined by the sum from n=1 to infinity of (x-11)^n / (n(-9)^n). The user applied the ratio test, arriving at the interval 2 < x < 20, which was confirmed as correct after checking the endpoints. The endpoints were verified, with x = 2 not converging and x = 20 converging, leading to the final conclusion that the interval of convergence is (2, 20].

PREREQUISITES
  • Understanding of power series and convergence
  • Familiarity with the ratio test for series convergence
  • Knowledge of endpoint convergence in power series
  • Basic algebra for simplifying inequalities
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  • Study endpoint convergence in power series
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Students studying calculus, particularly those focusing on series and sequences, as well as educators teaching convergence concepts in power series.

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Homework Statement



Find the interval of convergence for the given power series.
Sum from n=1 to infinty of (x-11)^n / (n(-9)^n)

Homework Equations





The Attempt at a Solution


I used the ratio test and I'm getting 2<x<20, but that doesn't seem to be right. I get abs(1/9*(x-11)) < 1, which simplifies into 2<x<20. I couldn't sworn I did this problem correctly, but I'm not getting the correct answer. Any help would be great.
 
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That is the correct interval, but don't forget you have to check the two end points separately.
 
Here the thing, I entered all possible compinations for the endpoints just to check my answer for the interval and they were all wrong. Just to double check it does not converge at 2 and does at 20, right?
 
Ok it accepts my answer now. It was probably something on my end but thanks anyways.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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