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.
On
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.
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.