Answer
$\dfrac{3}{4-x^2}$; $x \in (-2,2)$
Work Step by Step
We are given the power series $\Sigma_{k=0}^{\infty} (\dfrac{x^2-1}{3})^k$
As we know that $\Sigma_{k=0}^{\infty} x^k=\dfrac{1}{1-x}$
Therefore, the given series can be written as:
$\Sigma_{k=0}^{\infty} (\dfrac{x^2-1}{3})^k=\dfrac{1}{1-\dfrac{x^2-1}{3}}=\dfrac{3}{4-x^2}$
The given series will converge when $\dfrac{x^2-1}{3} \lt 1$
This implies that $|x^2-1| \lt 3 \implies -2 \lt x \lt 2$
So, our answers are: $\dfrac{3}{4-x^2}$; $x \in (-2,2)$
