Prove this sum does not converge uniformly I would like to prove that $\sum_{n=1}^\infty \frac{x^2}{(1+x^2)^n}$ converges for $x\in[-1,1]$, but not uniformly. Pointwise convergence is easy. 
Define $S_n(x)=\sum_{k=1}^n \frac{x^2}{(1+x^2)^k}$. 
$\frac{x^2}{(1+x^2)^n}\leq \left(\frac{x^2}{1+x^2}\right)^n\forall x\in[-1,1]$.
$\forall x\in[-1,1]\cap\{0\}$, $\sum_{n=1}^\infty\left(\frac{x^2}{1+x^2}\right)^n$ is a geometric series and converges to $\frac{1}{1-\frac{x^2}{1+x^2}}$. By the comparison test, $S_n(x)$ converges $\forall x\in[-1,1]\cap\{0\}.$ $S_n(0)=0 \forall n\in\mathbb{N}^*$.
Uniform convergence is proving much trickier. I've been trying to show that for some $x$ that is dependent on $n$, $S_n(x)$ is constant, which seems to be a commonly used trick to show something doesn't uniformly converge. I plotted part of the sum, and it looks like around $x=.05$, for sufficiently large $n$, $S_n(x)$ jumps from 0 to 1. Though I have no idea where I can pull this number out of the series.
 A: HINT:
Assume $x\in\mathbb{R}$:
$$\sum_{n=1}^{\infty}\frac{x^2}{\left(1+x^2\right)^n}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{x^2}{\left(1+x^2\right)^n}=\lim_{m\to\infty}\left(1-\frac{1}{\left(x^2+1\right)^m}\right)=$$
$$1-\lim_{m\to\infty}\frac{1}{\left(x^2+1\right)^m}=1-\frac{1}{\lim_{m\to\infty}\left(x^2+1\right)^m}=1-0=1$$
So we can say that:
$$\sum_{n=1}^{\infty}\frac{x^2}{\left(1+x^2\right)^n}\space\space\text{converges when}\space\frac{1}{|x^2+1|}<1$$
A: Define a sequence of functions $f_N: [-1,1] \rightarrow \mathbb{R}$, $N = 1, 2, 3, ...$ by $$f_N(x) = \sum\limits_{n=1}^{N} \frac{x^2}{(1+x^2)^n}$$ Then $f_N(x)$ converges pointwise to the function $f(x) = \sum\limits_{n=1}^{\infty} \frac{x^2}{(1+x^2)^n}$.  If $f_N$ converged uniformly to $f(x)$, then $f(x)$ would be a continuous function $[-1,1] \rightarrow \mathbb{R}$, since each function $f_N$ is continuous.
However, for $0 \neq x \in [-1,1]$, we have $$f(x) = x^2 \sum\limits_{n=1}^{\infty} \frac{1}{(1+x^2)^n} = \frac{x^2}{(1+x^2)} \frac{1}{(1- \frac{1}{1+x^2})} = \frac{x^2}{x^2} = 1$$  Here since $|\frac{1}{1+x^2}| < 1$ for $x \neq 0$, I can legally take $x^2$ out of the infinite sum, because the geometric series $\sum\limits_{n=1}^{\infty} \frac{1}{(1+x^2)^n}$ will converge on its own.  On the other hand, $f(0) = 0$.  So $f$ is not continuous. 
