Science:Math Exam Resources/Courses/MATH220/April 2011/Question 10 (b)
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Question 10 (b) 

Let be a bounded sequence of real numbers. For each define Prove that is a convergent sequence. (You may use part (a)). 
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 

Can you see why the sequence {b_{n}} is decreasing? How does that help show that the sequence is convergent? 
Hint 2 

To show that a decreasing sequence b_{n} converges, you have to show that it is bounded. Let's consider a few simpler examples to understand the relation between a_{n} and b_{n} better. E.g. consider the sequence or the other sequence Both sequences are bounded and satisfy Now in each case, consider the corresponding sequence b_{n} as describe in the question and observe. Note that those two cases are fairly easy as what will happen to that sequence (which you should be able to show converges). A more subtle case to consider after these two is the case of the sequence Notice that the first rational term converges to 1, but because of the sine term, that sequence will never converge. What will happen to the corresponding sequence b_{n} in this case? Why does b_{n} converge? 
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. The sequence {a_{n}} is bounded, which means that there exists a real number M such that Hence the supremum of all the numbers a_{n} is at most M and by definition of the numbers b_{n} we have that or in other words, the sequence {b_{n}} is bounded as well. That sequence will be converging if it is decreasing, that is if Which we can easily show to be true. Indeed, if we denote by then and clearly and by part (a) we have that This concludes our proof. Advanced note: We call the limit of this convergent sequence the limit superior of the original sequence {a_{n}}. 