91Ó°ÊÓ

$\underset{k=1}{\overset{∞}{∑}}\frac{1+\ln k}{k^2}$.

1. If $\underset{k→∞}{\lim }\frac{a_k}{b_k}=L$, where $L$is any positive real number, then either both converge or both diverge.
2. If $\underset{k→∞}{\lim }\frac{a_k}{b_k}=0$and $\underset{k=1}{\overset{∞}{∑}}{b}_k$converges, then role="math" localid="1649174943417" $\underset{k=1}{\overset{∞}{∑}}{a}_k$converges.
3. If$\underset{k→∞}{\lim }\frac{a_k}{b_k}=∞$and$\underset{k=1}{\overset{∞}{∑}}{b}_k$diverges, then$\underset{k=1}{\overset{∞}{∑}}{a}_k$diverges.

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

$1+\ln k≤\sqrt{k}$

Find the value of $\underset{k=1}{\overset{∞}{∑}}{b}_k$for the given series.

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

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

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

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

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

The value of $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$is convergent by $p-$series test.

Therefore, the value of $\underset{k=1}{\overset{∞}{∑}}{a}_k$is also convergent.

Hence, the given series is convergent.