Let $f_n(x) = \frac{x^2}{(1+x^2)^n}$ If $\sum_{n=0}^\infty$ converges pointwise, find it's limit. 
Let $f_n(x) = \frac{x^2}{(1+x^2)^n}$ If $\sum_{n=0}^\infty f_n(x)$ converges pointwise, find it's limit. Determine if it converges uniformly.

It is clear that $0\le \frac{x^2}{(1+x^2)^n} < 1, \forall n \ge 1$.
With the assumption is does converge pointwise it appears that it is a geometric series so we have.
$\displaystyle \sum_{n = 0}^\infty f_n(x) = \sum_{n = 0}^\infty \frac{x^2}{(1+x^2)^n} = \frac{x^2}{1-\frac{1}{(1+x^2)}}$
Being geometric is seems that it would converge uniformly but I am having difficulty showing why that is the case. 
 A: Note that$$\sum_{n=0}^N\frac{x^2}{(1+x^2)^n}=x^2\frac{1-\frac1{(1+x^2)^{N+1}}}{1-\frac1{1+x^2}}=1+x^2-\frac1{(1+x^2)^N}.$$So, your sum converges pointwise to $$\begin{array}{rccc}s\colon&\mathbb{R}&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}1+x^2&\text{ if }x\neq0\\0&\text{ if }x=0.\end{cases}\end{array}$$Since $s$ this is not a continuous functions, the convergence is not unform.
A: Note that we have 
$$\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}=x^2\left(\frac{1-\frac{1}{(1+x^2)^{N+1}}}{1-\frac{1}{1+x^2}}\right)=1+x^2-\frac{1}{(1+x^2)^N}$$
For $x>0$, we have $\lim_{N\to \infty }\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}=1+x^2$.

Now, take $\epsilon=\frac1e$.  Then, 
$$\left|\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}-\left(1+x^2\right)\right|=\frac{1}{(1+x^2)^N}$$
Take $x=\frac{1}{\sqrt N}$.  Then, note that 
$$\lim_{N\to\infty}\frac{1}{(1+x^2)^N}=\lim_{N\to\infty}\frac{1}{\left(1+\frac1N\right)}=\frac1e$$
Therefore, the convergence is not uniform on $(0,\infty)$, but is uniform on $[\delta,\infty)$ for any $\delta>0$.
