Show that $\sum\limits_{n=1}^{\infty}\frac{x}{1+n^2x^2}$ is not uniformly convergent in $[0,1]$ 
Show that $\sum\limits_{n=1}^{\infty}\frac{x}{1+n^2x^2}$ is not uniformly convergent in $[0,1]$.

I was thinking in the direction of taking the maximum value of each term $\frac{x}{1+n^2x^2}$, which is $\frac{1}{2n}$, and of summing them. That is clearly a divergent series.
But then those maximum values don't occur at the same value of $x$ for each term. For the $n$th term the maximum occurs at $x = \frac{1}{n}$.
So, how to proceed?
 A: You are on the right track. For clarity, let's denote the partial sum $\sum_{n = 1}^N \frac{x}{1 + n^2x^2}$ by $S_N(x)$. One way to show the result is to investigate $\sup_{x \in [0, 1]}|S_{2N}
(x) - S_{N}(x)|$: 
\begin{align}
\sup_{x \in [0, 1]} |S_{2N}(x) - S_N(x)| & = \sup_{x \in [0, 1]}\sum_{n = N + 1}^{2N} \frac{x}{1 + n^2x^2} \geq \sum_{n = N + 1}^{2N} \frac{1/N}{1 + n^2/N^2} \\
& \geq \frac{1}{N} \times N \times \frac{1}{1 + (2N)^2/N^2} = \frac{1}{5}
\end{align}
which doesn't converge to $0$ as $N \to \infty$. Thus we do not have uniform convergence (if $\{S_N(x)\}$ converges uniformly, then the above quantity is bounded to converge to $0$ as $N \to \infty$, in view of Cauchy's criterion).
A: The problem is that $$\lim_{x \rightarrow 0}\; \sum_{n=1}^{\infty} \frac{x}{1 + n^2 x^2}$$ is not $0$, so the function defined by the series isn't continuous. You can see that by evaluating at $x = 1/k:$ $$\sum_{n=1}^{\infty} \frac{1/k}{1 + n^2 / k^2}$$ is a Riemann sum for the integral $$\int_0^{\infty} \frac{1}{1+y^2} \, \mathrm{d}y$$ and so it tends to $\pi/2$ as $k \rightarrow \infty.$
