Finding a closed form for $\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$ So I got the sum
$$\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$$
and I want a closed form.
Can I just do
$$\frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1}{x^2} + 1 $$
for a closed form?

The formula that you are using for the sum of a geometric series, namely that
$$
\sum_{n=0}^\infty ar^n = \frac{a}{1-r},
$$
is valid only when $|r|<1$. In your case, the common ratio $r$ is
$$
\frac{1}{1+x^2},
$$
which is less than $1$ for all $x \neq 0$. So, for $x \neq 0$, you can express your series as
$$
\frac{\frac{1}{1+x^2}}{1-\frac{1}{1+x^2}} = \frac{1}{x^2}.
$$
Note that you missed the initial term $a = \frac{1}{1+x^2}$ in your computation.

When $x = 0$, the series diverges because it is the infinite sum $1 + 1 + 1 + \cdots$.
