Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (b)

MATH105 April 2012

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Question 08 (b)
Short-answer questions. No credit will be given for the answer (even if it is correct) without the accompanying work. (b) Find the limit, if it exists, of the sequence {ak}, where $a_{k}={\frac {k!\sin ^{3}(k)}{(k+1)!}}$

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
To start with, can you simplify the factorials?

Hint 2
After your simplification, what can you say about the numerator and the denominator separately, as $k\rightarrow \infty$?

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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. We wish to find the limit of the sequence $a_{k}={\frac {k!\sin ^{3}(k)}{(k+1)!}}$. Note: this is a sequence, not a series! The ratio test does not apply here, despite the fact we see factorials. Let's first simplify this a little. Note that $(k+1)!=(k+1)k!$, so really $a_{k}={\frac {k!\sin ^{3}(k)}{(k+1)k!}}={\frac {\sin ^{3}(k)}{k+1}}$ We now observe that the denominator goes to infinity for large values of k, while the enumerator remains bounded between -1 and +1. Mathematically, $-1\leq \sin ^{3}(k)\leq 1$. Hence, we expect the sequence to converge to zero. To prove this, we use the squeeze theorem: $\lim _{k\rightarrow \infty }{\frac {-1}{k+1}}\leq \lim _{k\rightarrow \infty }{\frac {\sin ^{3}(k)}{k+1}}\leq \lim _{k\rightarrow \infty }{\frac {1}{k+1}}$. The first and third limits are both 0, hence $0\leq \lim _{k\rightarrow \infty }{\frac {\sin ^{3}(k)}{k+1}}\leq 0$ so the squeeze theorem tells us $\lim _{k\rightarrow \infty }a_{k}=\lim _{k\rightarrow \infty }{\frac {\sin ^{3}(k)}{k+1}}=0$.

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