Q 19. - Calculus

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.

Step-by-Step Solution

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 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 ∞   b k

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

Step 3. Evaluating lim k → ∞   a k b k

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

Step 4. Result

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