Q. 29 - Calculus

Question

In Exercises $21-30$ use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.

$\sum _{k=1}^{\infty } \frac{1+\ln k}{k^3}$.

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty } \frac{1+\ln k}{k^3}$ is convergent.

Step 1 . Given information

$\sum _{k=1}^{\infty } \frac{1+\ln k}{k^3}$.

Step 2 . The limit comparison test states that for ∑ k = 1 ∞   a k and ∑ k = 1 ∞   b k be two series with positive terms then,

1. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=L$, where $L$ is any positive real number, then either both converge or both diverge.
2. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=0$, and $\sum _{k=1}^{\infty } b_k$ converges, then $\sum _{k=1}^{\infty } a_k$ converges.
3. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\infty$, and $\sum _{k=1}^{\infty } b_k$ diverges, then $\sum _{k=1}^{\infty } a_k$ diverges.

Step 3 . The terms of the series ∑ k = 1 ∞   1 + ln   k k 3 are positive.

The expression $1+\ln k$ satisfies the following inequality:

$1+\ln k\le k$.

Find $\sum _{k=1}^{\infty } b_k$ for the given series.

$$\begin{gathered} \sum _{k=1}^{\infty } b_k=\sum _{k=1}^{\infty } \frac{k}{k^3} \\ =\sum _{k=1}^{\infty } \frac{1}{k^2} \end{gathered}$$

Step 4 . Next find lim k → ∞   a k b k for the given series.

$$\begin{gathered} \underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\underset{k\to \infty }{\lim } \frac{{\displaystyle \frac{1+\ln k}{k^3}}}{{\displaystyle \frac{1}{k^2}}} \\ =\underset{k\to \infty }{\lim } \frac{1+\ln k}{k} \end{gathered}$$

                $=0$ [ using L'Hopital's rule]

Step 5 . From the obtained values,

The value of $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=0$ which is a finite number.

The value of $\sum _{k=1}^{\infty } b_k=\sum _{k=1}^{\infty } \frac{1}{k^2}$ is convergent by $p-$series test.

Therefore, the series $\sum _{k=1}^{\infty } a_k$ is also convergent.

Hence, the given series is convergent.