Uniform convergence of $ f(x) = \sum_n \frac{x}{e^{(nx-1)^2}} $ 
Examine point convergence, uniform convergence and almost uniform convergence on $\mathbb R$ of series:
  $$ f(x) = \sum_n \frac{x}{e^{(nx-1)^2}} $$

Point convergence is easy to show - it converges. 
Lets try uniform:
$$ \sup_{x \in D} \left|\frac{x}{e^{(nx-1)^2}} \right| \ge\frac{1}{n} \mbox{ and series $\sum 1/n$ doesn't converge}$$
    So we don't have almost uniform convergence and uniform convergence too. 
But I heard that this solution is wrong because in sup theorem we have implication only in one side. So how can we show uniform convergence (or show that it doesn't uniformly converge)?
 A: As Μάρκος Καραμέρης said, $e^x\geq1+x$ for all $x$.
Now pick some $a>0$ and look at uniform convergence on $S=(-\infty,-a]\cup[a,\infty)$. The remainder of our series is
$$
R(x;m)=\sum_{n>m}xe^{-(nx-1)^2}
$$
Using our estimates:
$$
|R(x;m)|\leq\sum_{n>m}|x|e^{-(nx-1)^2}\leq\sum_{n>m}\frac{|x|}{x^2}\frac{1}{1/x^2+(n-1/x)^2}=\sum_{n>m}\frac{1}{|x|}\frac{1}{1/x^2 +(n-1/x)^2}
$$
We have for all $x\in S$ and for all $n>1+1/a$:
$$
\frac{1}{|x|}\frac{1}{1/x^2+(n-1/x)^2}\leq\frac{1}{a}\frac{1}{(n-1/a)^2}
$$
This last series converges by the comparison test with $1/n^2$, so the original series converges uniformly on $S$ as defined for each $a$. 
The series, however, does not converge uniformly on $\mathbb{R}$ as the limit function is discontinuous at $0$: first, we have $f(0)=0$. Now pick some large $m\in\mathbb{N}$ and look at the series for $f(1/m)$:
$$
f\left(\frac{1}{m}\right)=\frac{1}{m}\sum_{n=1}^{\infty}\exp\left(-\left(\frac{n}{m}-1\right)^2\right)
$$
For $n\in[m/2, 3m/2]$, we have that $|n/m-1|\geq1/2$ and there are more than, say, $m/10$ such terms (the precise amount is irrelevant). Thus we have a bound:
$$
f\left(\frac{1}{m}\right)>\frac{1}{m}\frac{m}{10}e^{-1/4}=\frac{1}{10}e^{-1/4}
$$
As the sequence $f(1/m)$ would converge to $f(0)=0$ if $f$ were continuous at $0$, but clearly it does not, $f$ must be discontinuous at $0$ and so cannot converge uniformly on any interval that contains $0$.
