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

MATH101 April 2013

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[hide] 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.

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[show] 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.

[show] Hint 2
Try to prove convergence.

[show] Hint 3
[Carefully] Use the squeeze theorem!

[show] Hint 4
For an alternative solution, what is the standard way to show convergence for an alternating series?

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[show] 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.

[show] 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 a_n 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 a_n} converges. This, however, follow as well because if a series $\sum _{n=1}^{\infty }b_{n}$ converges, then the sequence {b_n} 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.