Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)/Solution 1

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 ${\frac {1}{k^{7/6}}}$.

${\begin{aligned}\displaystyle \lim _{k\rightarrow \infty }{\frac {b_{k}}{a_{k}}}&=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{k^{4}+1}}{\sqrt {k^{5}+9}}}=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})\cdot k^{4}}}{\sqrt {(1+9k^{-5})\cdot k^{5}}}}\\&=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}{\frac {k^{4/3}}{k^{5/2}}}=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}k^{4/3-5/2}\\&=\lim _{k\rightarrow \infty }k^{7/6}{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}k^{-7/6}=\lim _{k\rightarrow \infty }{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}\\&=\lim _{k\rightarrow \infty }{\frac {\sqrt[{3}]{(1+k^{-4})}}{\sqrt {(1+9k^{-5})}}}={\frac {1}{1}}=1.\\\end{aligned}}$