<s>[INST] Does $\sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$ converge uniformly on $(-\infty,\infty)$? 
Does $\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
  converge uniformly on $(-\infty,\infty)$?

My attempt:
No. Consider the case where $x=0$, then 
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = 0$.
For $x \neq 0$, observe $\displaystyle 0 \lt \frac{1}{(1+x^2)^n} \lt 1$, so by geometric series formula
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
$\displaystyle = \frac{x^2}{1 - \frac{1}{1+x^2}} = 1+x^2$
(1) So clearly the series doesn't even converge for all $x$, let alone converge uniformly.
Now, my question is about the case where $x \neq 0$. Does it converge uniformly to $1 + x^2$?
(2) I think, "yes". By Dini's theorem for series the convergence of the series to $1 + x^2$ must be uniform since $1+x^2$ is continuous and $(-\infty,0) \cup (0,\infty)$ is compact.
Is my reasoning for (1) and (2) correct?
 [/INST] For (1) you got the idea but the argument is not exactly right. 
Indeed the series does converges for all $x$, as you show 
$$ \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = \begin{cases} 1+ x^2 & \text{if } x\neq 0,\\ 0 & \text{if }x=0.\end{cases}.$$
However it does not converges to a continuous function. Thus the convergence is not uniform (I am using that if $f_n \to f$ uniformly and $f_n$ are continuous for all $n$, then so if $f$. 
For (2), it does not converges uniformly to $1+x^2$ on $(-\infty, 0) \cup (0,\infty)$. The set $(-\infty, 0) \cup (0,\infty)$ is not compact. 
To check that the convergence is not uniform, write $f_k (x) = \sum_{n=0}^k \frac{x^2}{(1+x^2)^n}$. Then
$$\left| f(x) -  f_k(x)\right| = \left| 1+x^2 - x^2 \frac{1-(1+x^2)^{-(k+1)}}{1-(1+x^2)^{-1}}\right| = \frac{1}{(1+x^2)^k}. $$
Thus 
$$\sup_{x\neq 0} |f(x) - f_k(x)| = 1$$
for all $k$ and so the convergence is not uniform. 
</s>