Example 4.23. Investigate the convergence of the functional series

$$
\sum_{n=1}^{\infty}\left(\frac{x-2}{1-3 x}\right)^{n}
$$
Solution. Apply Theorem 4.19:

$$
\lim _{n \rightarrow \infty}\left|\frac{x-2}{1-3 x}\right|^{n+1}:\left|\frac{x-2}{1-3 x}\right|^{n}=\left|\frac{x-2}{1-3 x}\right|
$$

Next, solve the inequality

$$
\begin{gathered}
\left|\frac{x-2}{1-3 x}\right|<1
\end{gathered}
$$

Thus, the domain of absolute convergence of the series is

$$
D=\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{3}{4},+\infty\right)
$$

If $x \in \bar{D}=\left(-\frac{1}{2}, \frac{3}{4}\right)$, the series will be divergent. Investigate the series at the points $x=-\frac{1}{2}$ and $x=\frac{3}{4}$. When $x=-\frac{1}{2}$, we get the numerical series $\sum_{n=1}^{\infty}(-1)^{n}$, which diverges (the general term does not tend to 0). When $x=\frac{3}{4}$, we get the series $\sum_{n=1}^{\infty} 1^{n}=\infty$. Therefore, the series $\sum_{n=1}^{\infty}\left(\frac{x-2}{1-3 x}\right)^{n}$ converges absolutely if $x \in D=\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{3}{4},+\infty\right)$ and diverges if $x \in\left[-\frac{1}{2}, \frac{3}{4}\right]$.