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

MATH220 April 2011

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Question 10 (b)
Let $\{a_{n}\}_{n=1}^{\infty }$ be a bounded sequence of real numbers. For each $n\in \mathbb {N}$ define $b_{n}=\sup\{a_{m}\mid m\in \mathbb {N} {\text{ s.t. }}m\geq n\}.$ Prove that $\{b_{n}\}_{n=1}^{\infty }$ 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 {bn} is decreasing? How does that help show that the sequence is convergent?

Hint 2

To show that a decreasing sequence bn converges, you have to show that it is bounded. Let's consider a few simpler examples to understand the relation between an and bn better. E.g. consider the sequence $a_{n}=(-1)^{n}\quad {\text{for all }}n\geq 1$ or the other sequence $a_{n}=\sin \left({\frac {n}{10}}\right)\quad {\text{for all }}n\geq 1$ Both sequences are bounded and satisfy $-1\leq a_{n}\leq 1\quad {\text{for all }}n\geq 1$ Now in each case, consider the corresponding sequence bn 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 $a_{n}={\frac {n+10}{n}}\sin \left({\frac {n}{10}}\right)\quad {\text{for all }}n\geq 1$ 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 bn in this case? Why does bn 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. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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 sequence {an} is bounded, which means that there exists a real number M such that $\displaystyle |a_{n}|\leq M\quad {\text{for all }}n\geq 1$ Hence the supremum of all the numbers an is at most M and by definition of the numbers bn we have that $\displaystyle |b_{n}|\leq M\quad {\text{for all }}n\geq 1$ or in other words, the sequence {bn} is bounded as well. That sequence will be converging if it is decreasing, that is if $\displaystyle b_{n+1}\leq b_{n}\quad {\text{for all }}n\geq 1$ Which we can easily show to be true. Indeed, if we denote by $\displaystyle B_{n}=\{a_{m}\mid m\in \mathbb {N} {\text{ s.t. }}m\geq n\}\quad {\text{for all }}n\geq 1$ then $\displaystyle b_{n}=\sup(B_{n})\quad {\text{for all }}n\geq 1$ and clearly $\displaystyle B_{n+1}\subseteq B_{n}\quad {\text{for all }}n\geq 1$ and by part (a) we have that $\displaystyle b_{n+1}=\sup(B_{n+1})\leq \sup(B_{n})=b_{n}$ This concludes our proof. Advanced note: We call the limit of this convergent sequence the limit superior of the original sequence {an}.

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