Answer
$\dfrac{x-2}{11-x}$; $x \in (-7,11)$
Work Step by Step
We are given the power series $\Sigma_{k=0}^{\infty} \dfrac{(x-2)^k}{3^{2k}}$
As we know that $\Sigma_{k=0}^{\infty} x^k=\dfrac{1}{1-x}$
Therefore, the given series can be written as:
$\Sigma_{k=0}^{\infty} \dfrac{(x-2)^k}{3^{2k}}=\dfrac{1}{1-\dfrac{x-2}{9}}=\dfrac{9}{11-x}$
and $\Sigma_{k=1}^{\infty} \dfrac{(x-2)^k}{3^{2k}}=\dfrac{9}{11-x}-1=\dfrac{x-2}{11-x}$
The given series will converge when $\dfrac{x-2}{11-x} \lt 1$
This implies that $|x-2| \lt 9 \implies -7 \lt x \lt 11$
So, our answers are: $\dfrac{x-2}{11-x}$; $x \in (-7,11)$
