Q. 25 - 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.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

Step 2 . The 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 term of the series ∑ k = 1 ∞   1 + ln   k k are positive.

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

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

Next find the $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}$ for the given series.

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

Step 4 . From the obtained values,

The value of $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=1$ which is finite non zero number.

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

Therefore, $\sum _{k=1}^{\infty } a_k$ is divergent.

Th given series is divergent.