91Ó°ÊÓ

Let us assume $b_k=\frac{\ln k}{k^2}{\left(x+\frac{5}{3}\right)}^k$therefore

$b_{k+1}=\frac{\ln \left(k+1\right)}{{\left(k+1\right)}^2}{\left(x+\frac{5}{3}\right)}^{k+1}$

The ratio for the absolute convergence is

$$\begin{gathered} \underset{k→∞}{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k→∞}{\lim }\left|\frac{{\displaystyle \frac{\ln \left(k+1\right)}{{\left(k+1\right)}^2}{\left(x+\frac{5}{3}\right)}^{k+1}}}{{\displaystyle \frac{\ln k}{k^2}{\left(x+\frac{5}{3}\right)}^k}}\right| \\ =\underset{k→∞}{\lim }\left|{\left(\frac{k}{k+1}\right)}^2\frac{\ln \left(k+1\right)}{\ln k}\left(x+\frac{5}{3}\right)\right| \\ =\underset{k→∞}{\lim }\left|x+\frac{5}{3}\right|{\left(\frac{k}{k+1}\right)}^2\frac{\ln \left(k+1\right)}{\ln k} \end{gathered}$$

Here the limit $\left|x+\frac{5}{3}\right|$is So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x+\frac{5}{3}\right|<1$that is$-\frac{8}{3}<x<-\frac{2}{3}$.

Now, since the intervals are finite so we analyze the behavior of the series at the endpoints.

So, when $x=-\frac{8}{3}$

localid="1649436283094" $$\begin{gathered} \underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(x+\frac{5}{3}\right)}^k=\underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(-\frac{8}{3}+\frac{5}{3}\right)}^k \\ =\underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(-1\right)}^k \end{gathered}$$

The result is just a constant multiple of the harmonic series, which converges conditionally.

So, when $x=-\frac{2}{3}$

$$\begin{gathered} \underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(x+\frac{5}{3}\right)}^k=\underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(-\frac{2}{3}+\frac{5}{3}\right)}^k \\ =\underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(1\right)}^k \end{gathered}$$

The result is just a constant multiple series, which converges at a single point.

Therefore, the interval of convergence of the power series localid="1649436616936" $\underset{k=1}{\overset{∞}{∑}}\frac{\ln k}{k^2}{\left(x+\frac{5}{3}\right)}^k$is $\left[-\frac{8}{3},-\frac{2}{3}\right]$.