Real Analysis, Pointwise Convergence but not uniform convergence Question:
Consider the series $g_n(x)=\sum_{k=0}^n\cfrac{x^2}{(1+x^2)^k}$
Prove that the series converges pointwise to the function
$$g(x)=\begin{cases} 0 & \text{ if } x=0 \\
 1+x^2 & \text{ if } x \neq 0
\end{cases}$$ 
but the convergence is not uniform on any interval containing $0$ on its interior.
Help? Not even sure what the first step is on this one.
 A: I'm assuming you're defining $$g_n(x)=\sum_{k=0}^n \frac{x^2}{(1+x^2)^k}$$
It is not hard to see that if $x=0$; $g_n(x)=0$ for each $n$ whence $g_n\to 0$. If $x\neq 0$, we then have that
$$ \lim \;g_n(x)=\sum_{k=0}^\infty \frac{x^2}{(1+x^2)^k}$$
$$ =x^2\sum_{k=0}^\infty \left(\frac{1}{1+x^2}\right)^k$$
$$  =x^2\frac{1}{1-\frac{1}{1+x^2}}$$
$$  =x^2\frac{1+x^2}{1+x^2-1}$$
$$  ={1+x^2}$$
and all the manipulations are justified for $$\frac 1 {1+x^2}<1$$ for any $x\neq 0$.
Can you see why the convergence is not uniform? If it were, your function would be continuous on an interval containing the origin.
A: The property used by Peter Tamaroff is the best way to prove that we do not have uniform convergence. It can also be done directly from the definition.
Suppose that $x\gt 0$. Then the absolute value of the difference between the limit $\dfrac{1}{1+x^2}$ and the sum of the terms up to $\dfrac{x^2}{(1+x^2)^m}$ is
$$\sum_{k=m+1}^\infty \frac{x^2}{(1+x^2)^k}.\tag{$1$}$$
This is an infinite geometric series. The usual formula gives that the sum $(1)$ is equal to $\dfrac{1}{(1+x^2)^m}$.
Let $\epsilon=1/2$. We show that there is no $N$ such that if $m\ge N$, then $(1)$ is $\lt \epsilon$ for every positive $x$. More informally, we show that there is no $N$ that "works" for every positive $x$. 
Suppose to the contrary that there is such an $N$, and let $m=N$. Then $\dfrac{1}{(1+x^2)^N}\lt 1/2$. By algebraic manipulation, this is true precisely if 
$$|x|\gt \sqrt{2^{1/N}-1}.\tag{$2$}$$ 
From $(2)$, we see that there are non-zero $x$ for which the inequality does not hold, namely the positive $x$ that are $\le \sqrt{2^{1/N}-1}$.  This contradicts the hypothesis of uniform convergence.  
