Science:Math Exam Resources/Courses/MATH101/April 2013/Question 12 (c)
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Question 12 (c) 

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. Determine, with explanation, whether converges or diverges. 
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 

By part (b), we cannot use integral test. It's clear geometric series, ratio test, and alternating series test do not apply. What's left? 
Hint 2 

Try either limit comparison test, or the direct comparison test. 
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 1 

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Please rate my easiness! It's quick and helps everyone guide their studies. Limit comparison test finishes this problem fairly quickly. By inspection, the summand looks like for large n. Let's prove this rigorously. and finite. Here, we used and both the left and right sides limit to 0, so squeeze theorem gives . Since diverges by pseries, we conclude diverges as well by the limit comparison test. 
Solution 2 

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Please rate my easiness! It's quick and helps everyone guide their studies. We can prove this using the comparison test, in the same way as part (a), with replacing and sums instead of integrals. Notice that for large , the summands behaves like , so this will be our goal. First, since , we have Now, for , we have To prove this, notice if , then where in the second line we added to both sides, and in the third line we divided by . Now, since diverges by pseries, we have that diverges by the comparison test. 