Answer
Converges to $\dfrac{x^2}{x^2+1}$ for all $|\dfrac{1}{x^2}| \lt 1$ or, $|x| \gt 1$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
We have a convergent geometric series with first term, $a=1$ and common ratio $r =\dfrac{-1}{x^2}$
$S=\dfrac{1}{1-(\dfrac{-1}{x^2})}=\dfrac{1}{1+\dfrac{-1}{x^2}}=\dfrac{x^2}{x^2+1}$
Hence, the given series converges to $\dfrac{x^2}{x^2+1}$ for all $|\dfrac{1}{x^2}| \lt 1$ or, $|x| \gt 1$
