Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)

MATH105 April 2014

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Question 02 (a)
Determine whether the series $\displaystyle \sum _{k=1}^{\infty }{\frac {\sqrt[{3}]{k^{4}+1}}{\sqrt {k^{5}+9}}}$ converges.

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
When k is very large, the largest contribution to the numerator is effectively $k^{4/3}$ and the largest contribution to the denominator is effectively $k^{5/2}$.

Hint 2
Compare the series with $\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{7/6}}}$.

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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. Since $\displaystyle {\frac {\sqrt[{3}]{k^{4}+1}}{\sqrt {k^{5}+9}}}\approx {\frac {\sqrt[{3}]{k^{4}}}{\sqrt {k^{5}}}}={\frac {k^{4/3}}{k^{5/2}}}={\frac {1}{k^{5/2-4/3}}}={\frac {1}{k^{7/6}}}$, we want to compare $\displaystyle {\frac {\sqrt[{3}]{k^{4}+1}}{\sqrt {k^{5}+9}}}$ with $\displaystyle {\frac {1}{k^{7/6}}}$. Let $\displaystyle a_{k}={\frac {1}{k^{7/6}}}>0$ and $\displaystyle b_{k}={\frac {\sqrt[{3}]{k^{4}+1}}{\sqrt {k^{5}+9}}}>0$. Recall $\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{\alpha }}}$ converges if and only if $\alpha >1$. Hence $\displaystyle \sum _{k=1}^{\infty }a_{k}$ converges. On the other hand, by the Limit Comparison Test, if $\displaystyle \lim _{k\rightarrow \infty }{\frac {b_{k}}{a_{k}}}=L$ for some nonzero finite value of L and both series are positive then either both $\displaystyle \sum _{k=1}^{\infty }a_{k}$ and $\displaystyle \sum _{k=1}^{\infty }b_{k}$ converge or both diverge. Hence, if we can prove $\displaystyle \lim _{k\rightarrow \infty }{\frac {b_{k}}{a_{k}}}=1$, then we know that $\displaystyle \sum _{k=1}^{\infty }b_{k}$ converges. They key fact here is that we know the convergence properties of one of the two series we are comparing, namely the series with ${\frac {1}{k^{7/6}}}$. ${\begin{aligned}\displaystyle \lim _{k\rightarrow \infty }{\frac {b_{k}}{a_{k}}}&=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{k^{4}+1}}{\sqrt {k^{5}+9}}}=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})\cdot k^{4}}}{\sqrt {(1+9k^{-5})\cdot k^{5}}}}\\&=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}{\frac {k^{4/3}}{k^{5/2}}}=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}k^{4/3-5/2}\\&=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}k^{-7/6}=\lim _{k\rightarrow \infty }{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}\\&=\lim _{k\rightarrow \infty }{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}={\frac {1}{1}}=1.\\\end{aligned}}$

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