[tex]f(x)=\dfrac1{8+x}=\dfrac18\dfrac1{1-\left(-\frac x8\right)}[/tex]
Recall that
[tex]\dfrac1{1-x}=\displaystyle\sum_{n\ge0}x^n[/tex]
for [tex]|x|<1[/tex], so we can write [tex]f(x)[/tex] as the series
[tex]f(x)=\displaystyle\frac18\sum_{n\ge0}\left(-\frac x8\right)^n[/tex]
