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. 

The MacLaurin series for $\frac{1}{1-x}$ converges when $|x| < 1$. If we substitute $-x^2$ for $x$, then our radius of convergence is when $|-x^2| < 1 \implies |x^2| < 1 \implies -1 < x < 1$.
Hope this helps!
