Answer
$-1 \lt x \lt 5$
$ f(x)=\displaystyle \frac{15}{5-x},\quad -1 \lt x \lt 5$
Work Step by Step
We aim for terms of Th.9.6 (Convergence of a Geometric Series).
A geometric series with ratio $r$ diverges when $|r| \geq 1$.
If $0 \lt |r| \lt 1$, then the series converges to the sum
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$.
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The series $\displaystyle \sum_{n=0}^{\infty}5(\frac{x-2}{3})^{n} $
$a=5, r=\dfrac{x-2}{3} $, converges for
$|r| \lt 1$
$|\displaystyle \frac{x-2}{3}| \lt 1$
$|x-2| \lt 3$
$-3 \lt x-2 \lt 3\quad/+2$
$-1 \lt x \lt 5$
$f(x)=\displaystyle \sum_{n=0}^{\infty}5(\frac{x-2}{3})^{n} =\frac{5}{1-(\frac{x-2}{3})}$
$=\displaystyle \frac{5}{\frac{3-x+2}{3}} =\frac{15}{5-x},\quad -1 \lt x \lt 5$
