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

MATH101 April 2013

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Question 02 (a)

Short-Answer Questions. Question 1-3 are short-answer 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 $\left\{(-1)^{n}\sin \left({\frac {1}{n}}\right)\right\}$ 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. 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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, note that $-\sin \left({\frac {1}{n}}\right)\leq (-1)^{n}\sin \left({\frac {1}{n}}\right)\leq \sin \left({\frac {1}{n}}\right)$ Further, since $\lim _{n\to \infty }{\frac {1}{n}}=0$ and $\displaystyle \sin x$ is continuous at 0, $\lim _{n\to \infty }\sin \left({\frac {1}{n}}\right)=\sin \left(\lim _{n\to \infty }{\frac {1}{n}}\right)=\sin(0)=0$ Similarly, $\lim _{n\to \infty }-\sin \left({\frac {1}{n}}\right)=-0=0.$ By the squeeze theorem, $\lim _{n\to \infty }(-1)^{n}\sin \left({\frac {1}{n}}\right)$ 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 $\sum _{n=1}^{\infty }(-1)^{n}a_{n}$ converges if an is monotonically decreasing. In our case, $a_{n}=\sin \left({\frac {1}{n}}\right)$, and since ${\frac {1}{n}}$ monotonically decreases as n increases, and sin(x) is a monotone function for x in [0,1] we find that indeed $a_{n}=\sin \left({\frac {1}{n}}\right)$ is monotonically decreasing. Hence $\sum _{n=1}^{\infty }(-1)^{n}a_{n}$ converges. But the question didn't ask if the series converges, but if the sequence {(-1)n an} converges. This, however, follow as well because if a series $\sum _{n=1}^{\infty }b_{n}$ converges, then the sequence {bn} must converge to zero, that is $0=\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }(-1)^{n}\sin \left({\frac {1}{n}}\right)$ Hence, $(-1)^{n}\sin \left({\frac {1}{n}}\right)$ converges to 0.

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