Q 20. - Calculus

Question

In Example 1 of Section 7.4 we used the integral test to show that the series $\sum _{k=1}^{\infty } \frac{k}{e^{k^2}}$ converges. Use the limit comparison test with the series $\sum _{k=1}^{\infty } \frac{1}{e^k}$ to prove the same result.

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} \mathrm{is} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = 0 \\ \mathrm{The} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{b}}_{\mathrm{k}} = \sum _{\mathrm{k}=1}^{\infty } \frac{1}{{\mathrm{e}}^{\mathrm{k}}} \mathrm{is} \mathrm{convergent}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{a}}_{\mathrm{k} }\mathrm{is} \mathrm{also} \mathrm{convergent}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty }\frac{\mathrm{k}}{{\mathrm{e}}^{{\mathrm{k}}^2}} \mathrm{is} \mathrm{convergent}. \end{gathered}$$

Step 1. Given information is:

$\sum _{k=1}^{\infty } \frac{k}{e^{k^2}}$

Step 2. Finding the term ∑   k = 1 ∞   b k

$$\begin{gathered} \mathrm{The} \mathrm{terms} \mathrm{of} \mathrm{the} \mathrm{series} \sum _{k=1}^{\infty } \frac{k}{e^{k^2}} \mathrm{are} \mathrm{positive}. \\ \mathrm{The} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{b}}_{\mathrm{k} }\mathrm{for} \mathrm{the} \mathrm{series} \sum _{k=1}^{\infty } \frac{k}{e^{k^2}} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \sum _{\mathrm{k}=1}^{\infty } {\mathrm{b}}_{\mathrm{k}} = \sum _{\mathrm{k}=1}^{\infty } \frac{1}{e^k} \end{gathered}$$

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

$$\begin{gathered} \mathrm{The} \mathrm{ratio} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = \underset{\mathrm{k}\to \infty }{\lim } \frac{{\displaystyle \frac{\mathrm{k}}{{\mathrm{e}}^{{\mathrm{k}}^2}}}}{{\displaystyle \frac{1}{{\mathrm{e}}^{\mathrm{k}}}}} \left(\mathrm{Substitution}\right) \\ =\underset{\mathrm{k}\to \infty }{\lim } \frac{\mathrm{k}}{{\mathrm{e}}^{\mathrm{k}}} \left(\mathrm{Simplify}\right) \\ =\underset{\mathrm{k}\to \infty }{\lim } \frac{1}{{\mathrm{e}}^{\mathrm{k}}} (\mathrm{Using} \mathrm{L}\text{'}\mathrm{Hopital}\text{'}\mathrm{s} \mathrm{rule}) \\ = 0 \end{gathered}$$

Step 4. Result

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} \mathrm{is} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = 0 \\ \mathrm{The} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{b}}_{\mathrm{k}} = \sum _{\mathrm{k}=1}^{\infty } \frac{1}{{\mathrm{e}}^{\mathrm{k}}} \mathrm{is} \mathrm{convergent}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{a}}_{\mathrm{k} }\mathrm{is} \mathrm{also} \mathrm{convergent}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty }\frac{\mathrm{k}}{{\mathrm{e}}^{{\mathrm{k}}^2}} \mathrm{is} \mathrm{convergent}. \end{gathered}$$