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.

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. The comparison test states that for ∑k=1∞ ak and ∑k=1∞ bk 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 role="math" localid="1649174943417" $\sum _{k=1}^{\infty }{a}_k$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 kk2 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.

localid="1649175485517" $$\begin{gathered} \sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2} \\ =\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

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

localid="1649175397761" $$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{1+\ln k}{k^2}}}{{\displaystyle \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ =\underset{k\to \infty }{\lim }\frac{1+\ln k}{k^{2-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\underset{k\to \infty }{\lim }\frac{1+\ln k}{k^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

$=0$[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.