Science:Math Exam Resources/Courses/MATH101/April 2013/Question 02 (a)
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Question 02 (a) 

ShortAnswer Questions. Question 13 are shortanswer questions. Put your answer in the box provided. Simplify your answer as much as possible. Full Marks will be awarded for a correct answer placed in the box. Show your work, for part marks. Consider the sequence State whether this sequence converges or diverges, and if it converges give its limit. 
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 

It might be difficult to figure out the exact values of this sequence. Instead, try using a theorem to prove convergence or divergence. 
Hint 2 

Try to prove convergence. 
Hint 3 

[Carefully] Use the squeeze theorem! 
Hint 4 

For an alternative solution, what is the standard way to show convergence for an alternating series? 
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. First, note that Further, since and is continuous at 0, Similarly, By the squeeze theorem, exists and equals 0 as well. 
Solution 2 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. The alternating series test says that converges if a_{n} is monotonically decreasing. In our case, , and since monotonically decreases as n increases, and sin(x) is a monotone function for x in [0,1] we find that indeed is monotonically decreasing. Hence converges. But the question didn't ask if the series converges, but if the sequence {(1)^{n} a_{n}} converges. This, however, follow as well because if a series converges, then the sequence {b_{n}} must converge to zero, that is Hence, converges to 0. 