Why doesn't this series converge uniformly? Could you tell me why the series $\displaystyle{\sum _k \frac{1}{1+k^2x^2}}$ doesn't converge uniformly on $(0,1]$?
 A: Because, if $x\rightarrow 0$ for fixed $k$, the individual terms tend to $1$. 
For uniform convergence you need to have that, independently of the choice of $x$, for each  $\varepsilon >0$ there is $N\in \mathbb{N}$ such that $|\sum_{k=n}^{m} a_k(x) | < \varepsilon$ for $n,m \ge N $. Suppose such an $N$ exists. Now choose $\varepsilon = 1/2$ and look at
$$ \sum_{N}^{N}a_k = \frac{1}{1+N^2 x^2}.$$
If $x\rightarrow 0$ this tends to $1 > \varepsilon$.
A: A necessary condition to establish the uniform convergence of a series of functions is for its general term, a sequence of functions, to converge uniformly.
Let's let $\:E=(0,1];\:f_n(x)=1/\left(1+n^2x^2\right).$
We have $$\forall x\in E\:\:\forall\varepsilon>0\:\:\exists N_{x,\varepsilon}>0\:\:\forall n\geqslant N_{x,\varepsilon}\implies|f_n(x)-0|<\varepsilon,\\\text{where }\:N_{x,\varepsilon}>\frac{\sqrt{1-\varepsilon}}{x}.$$
Although, $$\exists\varepsilon\in(0,1/2]\:\:\forall N_{\varepsilon}>0\:\:\exists x_n=\frac{1}n\in E\:\:\:\exists n\geqslant N_{\varepsilon}\implies|f_n(1/n)-0|\geqslant\varepsilon\\$$
$$\implies f_n\overset{\text{unif.}}{\not\to}f\equiv 0.$$
