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