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. 
 A: 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!
A: Hint: For which real numbers $x$ is 
$$
-1 < x^2 < 1 ?
$$
That's where you went off the rails. 
Don't try to apply a formula or anything -- just look at the graph of $y = x^2$. 
A: 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 $x<1.$
A: The series representation for $f(x)=\frac1{1+x^2}$ is given by
$$f(x)=\sum_{n=0}^\infty (-1)^nx^{2n}$$
for $|x|<1$, which can be shown using the root test, for example (i.e., $\limsup_{n\to \infty}\sqrt[n]{\left|(-1)^nx^{2n}\right|}=x^2$ is less than $1$ for $|x|<1$, and is greater than $1$ for $|x|>1$).  At the end points, we see trivially that the series diverges.  

Alternatively, we can write
$$f_N(x)=\sum_{n=0}^N(-1)^nx^{2n}=\frac{1-(-x^2)^{N+1}}{1+x^2}$$
For $|x|<1$, it is straightforward to show that $\lim_{N\to \infty}f_N(x)=\frac{1}{1+x^2}$, while for $|x|\ge 1$, the sequence $f_N(x)$ diverges.
