Uniform convergence: $\sum \frac{x^2}{(1 + x^2)^n}$ Consider the series of functions: 
$$\sum_{n \ge 1} \frac{x^2}{(1 + x^2)^n}$$

Q: Where does this series converge uniformly?

We have if $x \neq 0$:
$$\lim_{n \to \infty} \left| \frac{x^2}{(1 + x^2)^{n+1}} / \frac{x^2}{(1 + x^2)^n} \right|= \frac{1}{1 + x^2} < 1$$
And if $x = 0$, then the series converges. Therefore, we have pointwise convergence everywhere. 
For uniform convergence: 
$$R_n(x) = \sum_{k = n+1}^{\infty} \frac{x^2}{(1 + x^2)^k} \ge \sum_{k = n+1}^{2n} \frac{x^2}{(1 + x^2)^k} \ge \sum_{k = n+1}^{2n} \frac{x^2}{(1 + x^2)^{2n}} = \frac{nx^2}{(1 + x^2)^{2n}}, \ \forall x \in \mathbb {R}$$
Then:
$$R_n(\frac{1}{\sqrt{n}}) \ge \frac{1}{(1 + 1/n)^n)^2} \longrightarrow \frac{1}{e^2} > 0$$
So, we can't have uniform convergence in a neighbourhood of $0$.
It seems that for $a > 0$, we don't have uniform convergence on $[a, \infty)$ or $(-\infty, -a]$ because we can make $x$ as large as we desire. But, I'm not sure about this.
However, on any compact interval $[a,b]$ with $b < -1$ or $a > 1$, we do have uniform convergence because of normal convergence: 
$$\sup_{x \in [a,b]} \left( \frac{x^2}{(1 + x^2)^n} \right) \le \frac{b^2}{a^{2n}}$$
Can someone help me find all the sets on which we have uniform convergence? 
Thank you. 

Perhaps I am missing something, but I believe we can write your sequence as:
$$
\frac{x^2}{(1 + x^2)^n} = \frac{x^2 + 1 - 1}{(1 + x^2)^n} = \frac{1 + x^2}{(1 + x^2)^n} - \frac{1}{(1 + x^2)^n} = \frac{1}{(1 + x^2)^{n - 1}} - \frac{1}{(1 + x^2)^n}
$$
This gives two infinite sums:
\begin{align}
\sum_1^\infty \frac{x^2}{(1 + x^2)^n} =& \sum_1^\infty \left(\frac{1}{1 + x^2}\right)^{n - 1} - \sum_1^\infty \left(\frac{1}{1 + x^2}\right)^n \\
=&\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n - \left(\frac{1}{1 + x^2}\right)\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \\
=& \left(1 - \left(\frac{1}{1 + x^2}\right)\right)\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \\
=& \frac{x^2}{1 + x^2} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n
\end{align}
Since the only sum is now a geometric series with $r = \frac{1}{1 + x^2}$ giving the sum (when $|x| > 0$):
\begin{align}
\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n = \frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1 + x^2}{x^2} \\
\sum_1^\infty \frac{x^2}{(1 + x^2)^n} = \frac{x^2}{1 + x^2} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n = \frac{x^2}{1 + x^2} \frac{1 + x^2}{x^2} = 1
\end{align}
This series is identically $1$ everywhere except at $x = 0$ when the sum is clearly $0$.  Therefore this function $f(x) = \sum_1^\infty \frac{x^2}{(1 + x^2)^n}$ is discontinuous at the value $x = 0$ but continuous and differentiable everywhere else.
Another way to see the above is that the original rearrangement gave a telescopic series:
$$
1 - \frac{1}{1 + x^2} + \frac{1}{1 + x^2} - \frac{1}{(1 + x^2)^2} + \frac{1}{(1 + x^2)^2} ... - \frac{1}{(1 + x^2)^n}
$$
If $|x| > 0$ then $1 + x^2 > 1$ and thus $\lim_{n \rightarrow \infty}\left(1 - \frac{1}{(1 + x^2)^n}\right) = 1$.  Even when $x = 0$, for all $n$ we have: $1 - \frac{1}{1^n} = 0$ (which agrees with the observation that the sum is $0$ when $x = 0$).
