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
Formula to calculate the sum of a geometric series is:
$S=\dfrac{a}{1-r}$;
Here, $a=1$ and common ratio $r =\dfrac{-1}{x^2}$
Thus, $S=\dfrac{1}{1-(\dfrac{-1}{x^2})}$
or, $=\dfrac{1}{1+\dfrac{-1}{x^2}}$
or, $=\dfrac{x^2}{x^2+1}$
Hence, the series converges to $\dfrac{x^2}{x^2+1}$ for all $|\dfrac{1}{x^2}| \lt 1$ or, $|x| \gt 1$