Question

In Example 1 we used the comparison test to show that the series$\sum _{k=1}^{\infty }\frac{k^{\frac{3}{2}}-k-1}{5k^3+3}$ converges. Use the limit comparison test to prove the same result.

Short answer

$$\begin{gathered} \mathrm{The}\mathrm{value}\mathrm{of}\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}=\frac{1}{5};\mathrm{which}\mathrm{is}\mathrm{non}\mathrm{zero}\mathrm{finite}\mathrm{number}. \\ \mathrm{The}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\infty }\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}}\mathrm{is}\mathrm{convergent}\mathrm{by}\mathrm{p}-\mathrm{series}\mathrm{test}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{is}\mathrm{also}\mathrm{convergent}. \\ \mathrm{Hence},\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\frac{{\mathrm{k}}^{\frac{3}{2}}-\mathrm{k}-1}{5{\mathrm{k}}^3+3}\mathrm{is}\mathrm{convergent}. \end{gathered}$$

Step-by-step solution

Step 1. Given information is:

$\sum _{k=1}^{\infty }\frac{k^{\frac{3}{2}}-k-1}{5k^3+3}$

Step 2. Finding the term ∑k=1∞ bk

$$\begin{gathered} \mathrm{The}\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\frac{{\mathrm{k}}^{\frac{3}{2}}-\mathrm{k}-1}{5{\mathrm{k}}^3+3}\mathrm{are}\mathrm{positive}. \\ \mathrm{The}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}\mathrm{for}\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\frac{{\mathrm{k}}^{\frac{3}{2}}-\mathrm{k}-1}{5{\mathrm{k}}^3+3}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\infty }\frac{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}}{{\mathrm{k}}^3}(\mathrm{Dominant}\mathrm{term}\mathrm{of}\mathrm{numerator}\mathrm{and}\mathrm{denominator}) \\ =\sum _{\mathrm{k}=1}^{\infty }\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Step 3. Evaluating limk→∞ akbk

$$\begin{gathered} \mathrm{The}\mathrm{ratio}\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}=\underset{\mathrm{k}\to \infty }{\lim }\frac{\frac{{\mathrm{k}}^{\frac{3}{2}}-\mathrm{k}-1}{5{\mathrm{k}}^3+3}}{{\displaystyle \frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}}}}\left(\mathrm{Substitution}\right) \\ =\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}({\mathrm{k}}^{\frac{3}{2}}-\mathrm{k}-1)}{5{\mathrm{k}}^3+3}\left(\mathrm{Simplify}\right) \\ =\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{k}}^3\left(1-\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{1}{2}}}}-\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}}\right)}{{\mathrm{k}}^3\left(5+{\displaystyle \frac{3}{{\mathrm{k}}^3}}\right)} \\ =\underset{\mathrm{k}\to \infty }{\lim }\frac{\cancel{{\mathrm{k}}^3}\left(1-\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{1}{2}}}}-\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}}\right)}{\cancel{{\mathrm{k}}^3}\left(5+{\displaystyle \frac{3}{{\mathrm{k}}^3}}\right)}(\mathrm{cancel}\mathrm{out}\mathrm{common}\mathrm{factor}) \\ =\frac{1}{5} \end{gathered}$$

Step 4. Result

$$\begin{gathered} \mathrm{The}\mathrm{value}\mathrm{of}\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}=\frac{1}{5};\mathrm{which}\mathrm{is}\mathrm{non}\mathrm{zero}\mathrm{finite}\mathrm{number}. \\ \mathrm{The}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\infty }\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}}\mathrm{is}\mathrm{convergent}\mathrm{by}\mathrm{p}-\mathrm{series}\mathrm{test}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{is}\mathrm{also}\mathrm{convergent}. \\ \mathrm{Hence},\mathrm{the}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\frac{{\mathrm{k}}^{\frac{3}{2}}-\mathrm{k}-1}{5{\mathrm{k}}^3+3}\mathrm{is}\mathrm{convergent}. \end{gathered}$$