Uniform Convergence of $\sum_{n=1}^{\infty}e^{-nx^{2}}\sin(nx)$ in $[0,\infty)$ Consider the series $f(x)=\sum_{n=1}^{\infty}e^{-nx^{2}}\sin(nx)$:
a) Prove that this series converges uniformly on $[a,\infty)$, for each $a>0$
b) Does the series converge uniformly on $[0,\infty)?$
We can show uniform convergence on the interval $[a,\infty)$ with Weierstrass M-Test for part (a). For part (b), I suspect that there's a discontinuity at $x=0$ but I don't know how to justify it mathematically.
 A: For the series to converge uniformly on 
$[0,∞)$, given any $\epsilon > 0$  we must be able to find $N$ such that
$$\tag{*}\left|\sum_{k=n+1}^{m}e^{-kx^2}\sin kx\right| < \epsilon$$
for all $m >n > N$ and all $x \in [0,\infty)$ -- the uniform Cauchy criterion.  
However if we take $m = 2n$ and $x_n = \pi/(4n)$, then for $n <k < 2n$ we have $\sin kx_n > 1/\sqrt{2}$ and 
$$\left|\sum_{k=n+1}^{2n}e^{-kx_n^2}\sin kx_n\right|  > n\frac{1}{\sqrt{2}}e^{-2nx_n^2 } = \frac{ne^{-\pi^2/(8n)}}{\sqrt{2}} $$
where the RHS does not converge to $0$ as $n \to \infty$.
Hence, convergence is not uniform on $[0,\infty)$. Notice that $x_n = \pi/(4n) \to 0$ as $n \to \infty$ and 
non-uniform convergence arises for this series if $x$ can be arbitrarily close to $0$. 
A: Useful fact: If $\sum f_n$ converges uniformly on a set $E,$ then $\sup_E |f_n| \to 0$ as $n\to \infty.$ In our case we have
$$\sup_{[0,\infty)}|f_n| \ge f_n(1/n) = e^{-1/n}\sin (1) \to \sin (1)>0.$$
Therefore the given series does not converge uniformly on $[0,\infty).$
A: We could proceed by argument about continuity. We could try to compute the sum as follows: for $x >0$ we have $|\mathrm e^{-x^2\pm \mathrm ix}| = \mathrm e^{-x^2} <1$, thus
\begin{align*}
S(x) &=\frac 1{2\mathrm i}\sum_1^\infty \mathrm e^{-nx^2}(\mathrm e^{\mathrm inx} - \mathrm e^{-\mathrm inx})\\
&=\frac 1{2\mathrm i}\sum_0^\infty \mathrm e^{-nx^2}(\mathrm e^{\mathrm inx} - \mathrm e^{-\mathrm inx})\\
&=\frac 1{2\mathrm i}\sum_0^\infty \mathrm e^{n(\mathrm ix-x^2) }- \mathrm e^{n (-\mathrm ix - x^2)}\\
&= \frac 1 {2\mathrm i}\left(\frac {1} {1 - \exp(\mathrm ix -x^2)}-\frac 1 {1-\exp(-\mathrm ix-x^2)}\right)\\
&= \frac 1{2\mathrm i}\cdot \frac {\mathrm e^{-x^2} (\mathrm e^{\mathrm ix} - \mathrm e^{-\mathrm ix})}{(1-\exp(\mathrm ix -x^2) )(1 -\exp(-\mathrm ix -x^2))}\\
&= \mathrm e^{-x^2}\sin(x)\cdot \frac 1{1 + \mathrm e^{-2x^2}-2\mathrm {Re}(\mathrm e^{\mathrm ix -x^2})}\\
&=\frac {\mathrm e^{-x^2}\sin(x)} {1+\mathrm e^{-2x^2}- 2\mathrm e^{-x^2}\cos(x)}\\
&= \frac {\sin(x)} {\mathrm e^{x^2} + \mathrm e^{-x^2} - 2\cos(x)}.
\end{align*}
Since
\begin{align*}
&\phantom{=}\mathrm e^{-x^2} + \mathrm e^{x^2}- 2\cos(x) \\
&= 1 + (-x^2) + \frac {x^4}2 + 1 + x^2 + \frac {x^4}2 - 2\left(1 - \frac {x^2}2 +\frac {x^4}{24}\right) + o(x^4) \\
&= x^2 + O(x^4)\quad [x \to 0^+],
\end{align*}
we conclude that
$$
S(0^+) = \lim_{x \to 0^+}\frac {\sin(x)}{x^2 +O(x^4)} = +\infty, 
$$
while
$$
S(0) = 0,
$$
thus $S$ is not continuous on $[0,+\infty)$. 
Now assume the series converge uniformly on $[0,+\infty)$, then $S \in \mathcal C[0,+\infty)$, but this does not hold from above. Therefore the series does not converge uniformly on $[0,+\infty)$. 
