The valid interval of the maclaurin series for $\frac{1}{1+x^2}$ The Maclaurin series for $\frac{1}{1-x}$ is $1 + x + x^2 + \ldots$ for $-1 < x < 1$. 
To find the Maclaurin series for $\frac{1}{1+x^2}$, I replace $x$ by $-x^2$.
The Maclaurin series for $\frac{1}{1+x^2} = 1 - x^2 + x^4 - \ldots$.
This is valid for $-1 < -x^2 < 1$ if I replace $x$ by $-x^2$. So if I multiply each side by $-1$, I get $-1 < x^2 < 1$. If I take the square roots, I get $i < |x| < 1$. And I am stuck. 
And my book says that this Maclaurin series is valid for $-1 < x < 1$ anyway. There is no further explanation. 
How can I derive $-1 < x < 1$ from $i < |x| < 1$? Please help. 

Taking square roots in $a<x^2<b$ to get $\sqrt a < |x| < \sqrt b$ is valid if $a\ge0$ and $b\ge0$.
But notice that the solution of $-4< x^2$ is $-\infty<x<\infty$ since every square of a real number is $>-4.$
Since the square of a real number is never negative, the inequality $-1<x^2<1$ is equivalent to $x^2<1.$
That is equivalent to $x^2-1<0$, which is $(x-1)(x+1)<0$. A product of two numbers is negative only if one of them is negative and the other is positive. Since the second factor is bigger than the first, it needs to be the one that is positive. So we have $x+1>0$ and $x-1<0$. Thus $x>-1$ and $x<1.$
