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

Since $\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}\approx \frac{\sqrt[3]{k^4}}{\sqrt{k^5}}=\frac{k^{4/3}}{k^{5/2}}=\frac{1}{k^{5/2-4/3}}=\frac{1}{k^{7/6}}$, we want to compare $\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}$ with $\frac{1}{k^{7/6}}$.

Let $a_k=\frac{1}{k^{7/6}}>0$ and $b_k=\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}>0$. Recall $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ converges if and only if $\alpha >1$. Hence $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges.

On the other hand, by the Limit Comparison Test, if $\underset{k\to \mathrm{\infty }}{\lim }\frac{b_k}{a_k}=L$ for some nonzero finite value of L and both series are positive then either both $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ and $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ converge or both diverge. Hence, if we can prove $\underset{k\to \mathrm{\infty }}{\lim }\frac{b_k}{a_k}=1$, then we know that $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ 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 $\frac{1}{k^{7/6}}$.

$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{\lim }\frac{b_k}{a_k}& =\underset{k\to \mathrm{\infty }}{\lim }{k}^{7/6}\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}=\underset{k\to \mathrm{\infty }}{\lim }{k}^{7/6}\frac{\sqrt[3]{(1+k^{-4})\cdot k^4}}{\sqrt{(1+9k^{-5})\cdot k^5}}\\ & =\underset{k\to \mathrm{\infty }}{\lim }{k}^{7/6}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}\frac{k^{4/3}}{k^{5/2}}=\underset{k\to \mathrm{\infty }}{\lim }{k}^{7/6}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}{k}^{4/3-5/2}\\ & =\underset{k\to \mathrm{\infty }}{\lim }{k}^{7/6}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}{k}^{-7/6}=\underset{k\to \mathrm{\infty }}{\lim }\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}\\ & =\underset{k\to \mathrm{\infty }}{\lim }\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}=\frac{1}{1}=1.\\ \end{array}$