<s>[INST] 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?
 [/INST] You should write the first few terms to understand its behavior:
$$\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\frac{x^2}{(1+x^2)^3}+\cdots=$$
$$x^2\left(1+\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}+\cdots\right)\stackrel{x\ne 0}=$$
$$x^2\cdot \frac{1}{1-\frac{1}{1+x^2}}=1+x^2.$$
</s>