I think I'm able to show show if $f_n(x)$ defined on $R$ by $f_n(x)=\frac{1}{n+n^2x}$ is simply but not uniformly converging.
$\lim_\limits{n\rightarrow+\infty}\frac{1}{n+n^2x}=\frac{1}{n^2}$ (but I'm not sure why... only intuition...)
Yet, $\sum\frac{1}{n^2}$ is converging as a geometrical series. Thus $\sum f_n(x)$ converges simply.
Then how to show that it is uniformly converging?
For a fixed $x>0$, $$ \sum\frac1{n+n^2x}\leq\sum\frac1{n^2x}=\frac1x\,\sum\frac1{n^2}. $$ As the right hand side converges, the original series converges for each $x>0$.
But the convergence is not uniform: if it were, the limit function would admit a continuous extension to $x=0$, and we would have $\lim_{x\to0}f_n(x)$ finite. But $\lim_{x\to0}f_n(x)=\infty$, so the convergence cannot be uniform on $(0,\delta)$ for any $\delta>0$.
As long as $x \neq 0$, then:
$$\frac{1}{n+n^2x} = \frac{1}{x}\frac{1}{\left(\frac{n}{x}+n^2\right)} \sim \frac{1}{x}\frac{1}{n^2},$$
when $n$ goes to infinity.
This implies that $$\sum f_n(x) \sim \sum \frac{1}{x}\frac{1}{n^2} = \frac{1}{x}\sum \frac{1}{n^2}$$
and then $\sum f_n(x)$ converges since $\sum \frac{1}{n^2}$ converges too.
If $x=0$, then:
$$f_n(x) = \frac{1}{n}.$$
In this case $\sum f_n(x)$ diverges as $\sum \frac{1}{n}$ diverges too.
Since for $x=0$ you don't have simple convergence, the you can't have uniform convergence.
Indeed, the limit function $$f(x) = \sum f_n(x) = \sum \frac{1}{x}\frac{1}{\left(\frac{n}{x}+n^2\right)} \sim \frac{\pi^2}{6x},$$
is clearly undefined for $x=0$.
