If [tex]f(x)=\ln(5-x)[/tex], then
[tex]f'(x)=-\dfrac1{5-x}[/tex]
[tex]f''(x)=-\dfrac1{(5-x)^2}[/tex]
[tex]f'''(x)=-\dfrac2{(5-x)^3}[/tex]
and so on, following the general pattern
[tex]f^{(n)}(x)=-\dfrac{(n-1)!}{(5-x)^n}[/tex]
Then the power series for [tex]f(x)[/tex] centered at [tex]x=0[/tex] is
[tex]\displaystyle f(0)+\sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n[/tex]
[tex]\displaystyle \ln5+\sum_{n=1}^\infty\frac{(n-1)!}{n!5^n}x^n[/tex]
[tex]\displaystyle \ln5+\sum_{n=1}^\infty\frac{x^n}{n5^n}[/tex]
The power series converges by the ratio test for
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{(n+1)5^{n+1}}}{\frac{x^n}{n5^n}}\right|=\frac{|x|}5\lim_{n\to\infty}\frac n{n+1}=\frac{|x|}5<1[/tex]
or [tex]|x|<5[/tex], so that the radius of convergence is 5.
