Example 5.21. Expand the function $f(x)=\ln (1-2 x)$ into a Taylor series.
Solution. Replacing $x$ with $-2x$ in the expansion (7), we get:

$$
\ln (1-2 x)=(-2 x)-\frac{(-2 x)^{2}}{2}+\frac{(-2 x)^{3}}{3}-\ldots+(-1)^{n-1} \frac{(-2 x)^{n}}{n}+\ldots,
$$

or

$$
\ln (1-2 x)=-2 x-\frac{2^{2}}{2} x^{2}-\frac{2^{3}}{3} x^{3}-\ldots-\frac{2^{n}}{n} x^{n}-\ldots
$$

The expansion (7) is valid in the interval $-1 < x \leq 1$, and the desired expansion is obtained by replacing $x$ with $-2x$; therefore, to find the interval of convergence of the resulting series, we need to solve the inequality

$$
-1 < -2 x \leq 1
$$

from which

$$
-\frac{1}{2} \leq x < \frac{1}{2}
$$