Answer
$1 \lt x \lt 9$ or, $x \in (1,9)$
Work Step by Step
We are given the power series $\Sigma_{k=0}^{\infty} (\sqrt x-2)^k$
As we know that $\Sigma_{k=0}^{\infty} x^k=\dfrac{1}{1-x}$
Therefore, $\Sigma_{k=0}^{\infty} (\sqrt x-2)^k=\dfrac{1}{1-(\sqrt x-2)}=\dfrac{1}{3-\sqrt x}$
The series will converge when $|\sqrt x-2| \lt 1$
This implies that $1 \lt x \lt 9$
So, we have $x \in (1,9)$
