Q. 27 - 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.

$\sum _{k=1}^{\infty } \frac{1+\ln k}{k^2}$.

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty } \frac{1+\ln k}{k^2}$ is convergent.

Step 1 . Given information

$\sum _{k=1}^{\infty } \frac{1+\ln k}{k^2}$.

Step 2 . The comparison test states that for ∑ k = 1 ∞   a k and ∑ k = 1 ∞   b k be two series with positive term 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 2 are positive.

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

$1+\ln k\le \sqrt{k}$

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

$$\begin{gathered} \sum _{k=1}^{\infty } b_k=\sum _{k=1}^{\infty } \frac{\sqrt{k}}{k^2} \\ =\sum _{k=1}^{\infty } \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

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

$$\begin{gathered} \underset{k\to \infty }{\lim } \frac{a_k}{b_k} =\underset{k\to \infty }{\lim } \frac{{\displaystyle \frac{1+\ln k}{k^2}}}{{\displaystyle \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ = \underset{k\to \infty }{\lim } \frac{1+\ln k}{k^{2-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\underset{k\to \infty }{\lim } \frac{1+\ln k}{k^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

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

Step 4 . From the obtained values,

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

The value of $\sum _{k=1}^{\infty } b_k=\sum _{k=1}^{\infty } \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ is convergent by $p-$series test.

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

Hence, the given series is convergent.