To find the power series representation of In(5 - x), we recall that the power series representation of a function of the form:
[tex] \frac{a}{1-r} =\sum\limits^\infty_{n=0}ar^n[/tex]
provided |r| < 1
Also recall that
[tex]\int{\frac{a}{1-r}\,dx} =\int{\sum\limits^\infty_{n=0}ar^n\,dx}[/tex]
Notice that
[tex]\ln{(5-x)}=-\int {\frac{1}{5-x}\,dx} [/tex]
To get the power series of
[tex]\frac{1}{5-x}= (\frac{1}{5}) \frac{1}{1- \frac{x}{5} } \\ =\sum\limits^\infty_{n=0}( \frac{1}{5}) (\frac{x}{5})^n=(\frac{1}{5}) (\frac{x}{5})+(\frac{1}{5}) (\frac{x^2}{25})+(\frac{1}{5}) (\frac{x^3}{125})+ . . . \\ =\frac{x}{25}+\frac{x^2}{125}+ \frac{x^3}{625}+ . . . [/tex]
Therefore, the power series representation of
[tex]\ln{(5-x)}=-\int {\frac{1}{5-x}\,dx} \\ =-\int{(\frac{x}{25}+\frac{x^2}{125}+ \frac{x^3}{625}+ . . . ),dx} \\ =C-\frac{x^2}{50}-\frac{x^3}{375}- \frac{x^4}{2500}- . . . [/tex]
When x = 0: C = ln 5
Therefore,
[tex]\ln{(5-x)}=\ln{5}-\frac{x^2}{50}-\frac{x^3}{375}- \frac{x^4}{2500}- . . . [/tex]
The radius of convergence is given by |r| < 1.
Here,
[tex]r= \frac{x}{5}[/tex]
Therefore, radius of convergence is
[tex]| \frac{x}{5}| \ \textless \ 1 \\ |x|\ \textless \ 5[/tex]
