I want to see if the following sum converges or diverges:
$$\sum_{n=2}^{\infty}{(3-x^2)^n}$$
We had the intuition that this was a geometric series, therefore we found out the value of $|3-x^2|<1$.
This gives us the solutions $-2<x<-\sqrt{2}$ and $\sqrt{2}<x<2$. These are the values of which the sum should converge. However Wolfram says the series diverges by the geometric test. Any hints?
Yes you are absolutely right
$$\sum_{n=2}^{\infty}{(3-x^2)^n}=\sum_{n=2}^{\infty}{r^n}$$
is a geometric series converging for $|r|<1$ to
$$\sum_{n=2}^{\infty}{r^n}=\frac1{1-r}-1-r=\frac{r^2}{1-r}=\frac{(x^2-3)^2}{x^2-2}$$