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
Remember that the series will converge when $|f(x)|=|\dfrac{x^2-1}{2}| \lt 1$
This implies that $x^2 -1\lt 2$
and $-2 \lt x^2-1 \lt 2$
This means that the interval of convergence is : $-\sqrt 3 \lt x \sqrt 3$
Next, we will find the sum of the given geometric series
$S=\dfrac{a}{1-r} \\=\dfrac{1}{1-\dfrac{x^2-1}{2}}
\\=\dfrac{2}{3-x^2}$
So, $\Sigma_{n=0}^{\infty} (\dfrac{x^2-1}{2})^n=\dfrac{2}{3-x^2}$
