Answer
Interval of Convergence: $ -\sqrt 3 \lt x \sqrt 3$
and $\Sigma_{n=0}^{\infty} (\dfrac{x^2-1}{2})^n=\dfrac{2}{3-x^2}$
Work Step by Step
The series will converge when $|f(x)|=|\dfrac{x^2-1}{2}| \lt 1$
So, $x^2 -1\lt 2$
$ \implies -2 \lt x^2-1 \lt 2$
$ \implies -\sqrt 3 \lt x \sqrt 3$
Now, the sum of the given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{x^2-1}{2}}=\dfrac{2}{3-x^2}$
Thus, the Interval of Convergence is:
$ -\sqrt 3 \lt x \sqrt 3$
and $\Sigma_{n=0}^{\infty} (\dfrac{x^2-1}{2})^n=\dfrac{2}{3-x^2}$
