Science:Math Exam Resources/Courses/MATH101/April 2013/Question 11
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Question 11 

FullSolution Problems. In questions 4–12, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Consider the power series , where is a real number. Find the interval of convergence of this series. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Start by finding the radius of convergence. 
Hint 2 

Use the Ratio Test to calculate the radius of convergence. 
Hint 3 

Remember to check the endpoints of the interval! 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. We begin by calculating the radius of convergence, using the Ratio Test: By the Ratio Test, the series converges if and diverges if . Now we need to check the endpoints. When , that is, , we have which diverges by the pseries test, since . When , that is, , we have This is an alternating series with Clearly, and . By the Alternating Series Test, converges. Therefore, the interval of convergence is . 