## Calculus (3rd Edition)

$$f(x)=\ln(1-5x)=-5 x-\frac{ (5x)^2}{2}-\frac{(5x)^3}{3}-\frac{(5x)^4}{4}+...$$ Convergent for the values $|x|\lt 1/5$ and $x=-1/5$.
By making use of Table 2, we have the Maclaurin series for $f(x)=\ln(1+x)$ as follows $$f(x)=\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$ Now by the comparison with the function $f(x)=\ln(1-5x)$, we have the Maclaurin series as follows $$f(x)=\ln(1-5x)=-5 x-\frac{ (5x)^2}{2}-\frac{(5x)^3}{3}-\frac{(5x)^4}{4}+...$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=\ln(1+x)$ is convergent for the values $|x|\lt 1$ and $x=1$. In our case, we have $|-5x|\lt 1$ ($|x|\lt 1/5$) and $-5x=1$ ($x=-1/5$). Hence the Maclaurin series for $f(x)=\ln(1-5x)$ is convergent for the values $|x|\lt 1/5$ and $x=-1/5$.