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

FullSolution Problem. Justify your answer and show all your work. Simplification of numerical answers is required.. Explain why you cannot conclude that diverges from part (a) and the Integral Test. 
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 

What hypotheses are required for the Integral Test? Which one fails? 
Hint 2 

If , then its clear it is positive, tends to 0, and agrees with the summands on the natural numbers. So prove it is never eventually decreasing. 
Hint 3 

It might be easiest to prove is never eventually decreasing by differentiating and using a differential calculus result. 
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. If , then its clear it is positive, tends to 0, and agrees with the summands on the natural numbers. So we should prove it is never eventually decreasing. Taking the derivative, We want to show that ƒ is never eventually decreasing. For this, it suffices to find a sequence x_{n} of values that grow with no limit, and for which ƒ'(x_{n}) is positive. How would be go about finding such a sequence? Let us look at the enumerator more carefully. Our best bet to make ƒ'(x_{n}) positive is to choose x_{n} such that the positive term is maximal. This is achieved when cos(x_{n}) = 1. In this case, sin(x_{n}) = 0 and we find that the enumerator becomes 2(1+x_{n}^{2})2x_{n}^{2} = 2 > 0. Hence, we choose for any integer n, and double check that indeed the numerator becomes: The denominator is always positive, so for these values of , so that is increasing at these values of . Since grows with no limit, is never eventually decreasing, and hence the integral test fails. 