Science:Math Exam Resources/Courses/MATH105/April 2012/Question 02 (b)

MATH105 April 2012

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Question 02 (b)
This problem contains three numerical series. For each of them, find out whether it converges or diverges. You should provide appropriate justification in order to receive credit. $\sum _{k=1}^{\infty }{\frac {k^{4}-2k^{3}+2}{k^{5}+k^{2}+k}}$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Compare the highest exponents of the numerator and the denominator to make an educated guess. Then confirm your guess using the limit comparison test.

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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. When we look at this series, we note that the leading power of the numerator is k4 and the leading power of the denominator is k5. Thus, for k very large, we would expect the terms in the series to look like k4/k5 = 1/k. We will use the limit comparison test with $a_{k}={\frac {k^{4}-2k^{3}+2}{k^{5}+k^{2}+k}}{\text{ and }}b_{k}={\frac {1}{k}}$ We compute ${\begin{aligned}\lim _{k\rightarrow \infty }{\frac {a_{k}}{b_{k}}}&=\lim _{k\rightarrow \infty }({\frac {k^{4}-2k^{3}+2}{k^{5}+k^{2}+k}})/({\frac {1}{k}})\\&=\lim _{k\rightarrow \infty }{\frac {k^{5}-2k^{4}+2k}{k^{5}+k^{2}+k}}\\&=\lim _{k\rightarrow \infty }{\frac {k^{5}/k^{5}-2k^{4}/k^{5}+2k/k^{5}}{k^{5}/k^{5}+k^{2}/k^{5}+k/k^{5}}}\\&=\lim _{k\rightarrow \infty }{\frac {1-2/k+2/k^{4}}{1+1/k^{3}+1/k^{4}}}\\&={\frac {1-0+0}{1+0+0}}=1\end{aligned}}$ As this limit is nonzero and finite, $\sum _{k=1}^{\infty }a_{k}$ converges if and only if $\sum _{k=1}^{\infty }b_{k}$ converges. The series $\sum _{k=1}^{\infty }b_{k}=\sum _{k=1}^{\infty }{\frac {1}{k}}$ diverges (it's worth remembering - or being able to show with the integral test - that $\sum _{k=1}^{\infty }{\frac {1}{k^{p}}}$ converges if $p>1$ and diverges if $p\leq 1$). The conclusion is that the series diverges!

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