The sum of an infinite geometric series is given by:
$$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$$
In this case, $a = 1$ and $r = -x^2$, so the sum is:
$$\frac{1}{1-(-x^2)} = \frac{1}{1+x^2}$$
So, we have shown that for $-1 < x < 1$,
$$\frac{1}{1+x^2} \approx
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