<s>[INST] 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?
 [/INST] Another way of solving it is by bringing it to the standard form, then you can use the formula for infinite geometric series considering $x \ne 0$
take $n=m+1$, so you have:
$$ \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m+1}}= \frac{1}{(1 + x^2)} \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m}}=\frac{1}{(1 + x^2)}\cdot \frac{(1 + x^2)}{ x^2}=\frac{1}{x^2}$$
</s>