Looks like the given function is
[tex]f(x)=\dfrac1{9-x}[/tex]
Recall that for |x| < 1, we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
We want the series to be centered around [tex]x=4[/tex], so first we rearrange f(x) :
[tex]\dfrac1{9-x}=\dfrac1{5-(x-4)}=\dfrac15\dfrac1{1-\frac{x-4}5}[/tex]
Then
[tex]\dfrac1{9-x}=\displaystyle\frac15\sum_{n=0}^\infty\left(\frac{x-4}5\right)^n[/tex]
which converges for |(x - 4)/5| < 1, or -1 < x < 9.
