Issue with Baby Rudin Example 7.3 Example 7.3 of Baby Rudin states that the sum
\begin{align}
\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}
\end{align}
is, for $x \neq 0$, a convergent geometric series with sum $1 + x^2$.  This confuses me.  As far as I know, a geometric series is a series of the form
\begin{align}
\sum_{n=0}^{\infty}x^n,
\end{align}
and such a series converges to $\frac{1}{1 - x}$ if $x \in [0,1)$.  Now, it is certainly true that if we put $y = \frac{x^2}{1 + x^2}$, then $\frac{1}{1 - y} = 1 + x^2$.  But in this case, shouldn't the series we're discussing be
\begin{align}
\sum_{n=0}^{\infty}\left[\frac{x^2}{1 + x^2}\right]^n?
\end{align}
What am I missing?

$$\begin{align}
\sum_{n=0}^\infty \frac{x^2}{(1+x^2)^n}&=x^2\sum_{n=0}^\infty \left((1+x^2)^{-1}\right)^n\\\\
&=\frac{x^2}{1-(1+x^2)^{-1}}\\\\
&=\frac{x^2(1+x^2)}{1+x^2-1}\\\\
&=1+x^2
\end{align}$$
for $x\ne0$.
