Science:Math Exam Resources/Courses/MATH103/April 2013/Question 01 (b)

MATH103 April 2013

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[hide] Question 01 (b)
Sequences and Series: Short Answer Problems - Determine whether the following sequences and series converge (full marks for correct answer with justification; work must be shown for partial marks) Does the sequence $\left({\frac {n^{3}+2}{{\sqrt {n^{6}+3}}+{\sqrt[{3}]{27n^{9}+1}}}}\right)_{n\geq 1}$ converge? If it does, calculate the 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.

[show] Hint 1
Don't let the square and cube roots fool you! This is still essentially a polynomial divided by a polynomial so factor out the largest power of n in both the numerator and the denominator.

[show] Hint 2
What are the dominant terms when $n$ is large?

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[show] Solution

Proceeding as in the hint, we have ${\begin{aligned}{\frac {n^{3}+2}{{\sqrt {n^{6}+3}}+{\sqrt[{3}]{27n^{9}+1}}}}&={\frac {n^{3}(1+{\tfrac {2}{n^{3}}})}{{\sqrt {n^{6}(1+{\tfrac {3}{n^{6}}})}}+{\sqrt[{3}]{27n^{9}(1+{\tfrac {1}{27n^{9}}})}}}}\\&={\frac {n^{3}(1+{\tfrac {2}{n^{3}}})}{n^{3}{\sqrt {(1+{\tfrac {3}{n^{6}}})}}+3n^{3}{\sqrt[{3}]{(1+{\tfrac {1}{27n^{9}}})}}}}\\&={\frac {1+{\tfrac {2}{n^{3}}}}{{\sqrt {(1+{\tfrac {3}{n^{6}}})}}+3{\sqrt[{3}]{(1+{\tfrac {1}{27n^{9}}})}}}}\end{aligned}}$ Taking the limit as n tends to infinity in the last equality above yields $\lim _{n\to \infty }{\frac {n^{3}+2}{{\sqrt {n^{6}+3}}+{\sqrt[{3}]{27n^{9}+1}}}}=\lim _{n\to \infty }{\frac {1+{\tfrac {2}{n^{3}}}}{{\sqrt {(1+{\tfrac {3}{n^{6}}})}}+3{\sqrt[{3}]{(1+{\tfrac {1}{27n^{9}}})}}}}={\frac {1}{1+3}}={\frac {1}{4}}$ and so the sequence converges to one fourth.

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