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