Answer
$|x^2+1| \geq 1$; diverges.
Work Step by Step
We are given the series $f(x)=\Sigma_{k=0}^{\infty} (x^2+1)^{2k}$
Since, we know that $\Sigma_{k=0}^{\infty} x^{k}$ is the power expansion of $f(x)=\dfrac{1}{1-x}$
Thus, we can write as;
$\Sigma_{k=0}^{\infty} (x^2+1)^{2k}=\dfrac{1}{1-(x^2+1)^2}=-\dfrac{1}{x^2(2+x^2)}$
Here, we can see that |x^2+1| \geq 1$. This implies that the series diverges.
