The Taylor Series expansion for $\frac{1}{1-x}$ is convergent for every real number $-1 < x < 1$.
\begin{equation*}\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^{n} .\end{equation*}
Since $0 \leq x^{2} < 1$ if, and only if $-1 < x < 1$, the Taylor Series expansion for $\frac{1}{1-x^{2}}$ is convergent for every real number $-1 < x < 1$.
\begin{equation*}\frac{1}{1 - x^{2}} = \sum_{n=0}^{\infty} x^{2n} .\end{equation*}
This is a special case of a theorem about the composition of certain functions. For example, the Taylor Series expansion for $e^{x}$ implies that
\begin{equation*}e^{x^{2}} = \sum_{n=0}^{\infty} \frac{1}{n!} \, x^{2n} \end{equation*}
and that
\begin{equation*}e^{(x-1)^2}= \sum_{n=0}^{\infty} \frac{1}{n!} \, \left((x - 1)^2\right)^{n}= \sum_{n=0}^{\infty} \frac{1}{n!} \, (x - 1)^{2n} .\end{equation*}
I am looking for the precise statement for such a theorem and a rigorous demonstration of it (or a citation for a rigorous demonstration of it).
