Uniform convergence on $[0,+\infty[$ of the series $\sum_{n=1}^{+\infty} \frac{\mathrm e^{-nx}\sin (nx)}{\sqrt n}$

I'am looking to study the uniform convergence on $[0,+\infty[$ of the series
$$\sum_{n=1}^{+\infty} \frac{\mathrm e^{-nx}\sin (nx)}{\sqrt n}$$ The series converges pointwise on $[0,+\infty[$, it also normally converges on $[a,+\infty[$, $\forall$ a>0 But I don't see how to prove or refute the uniform convergence on $[0,+\infty[$ , I tried with Abel's test for uniform convergence series without success: difficulty to prove that $\sum_{n=N}^{+\infty} \mathrm e^{-nx}\sin (nx)$ is bounded uniformly in $x\in [0,+\infty[$ Thank you in advance for help or answer

Assume that it were, then \begin{align*} \left|\sum_{k=n}^{2n}\dfrac{e^{-kx}\sin(kx)}{\sqrt{k}}\right|<1 \end{align*} for sufficiently large $n$ and $x\in[0,\infty)$.

Put $x=x_{n}=\pi/(4n)$ then $\sin(kx_{n})\geq 1/\sqrt{2}$ for all $n\leq k\leq 2n$, and hence \begin{align*} \left|\sum_{k=n}^{2n}\dfrac{e^{-kx_{n}}\sin(kx_{n})}{\sqrt{k}}\right|&=\sum_{k=n}^{2n}\dfrac{e^{-kx_{n}}\sin(kx_{n})}{\sqrt{k}}\\ &\geq\dfrac{1}{\sqrt{2}}\sum_{k=n}^{2n}\dfrac{e^{-2n\cdot\pi/(4n)}}{\sqrt{k}}\\ &=\dfrac{e^{-\pi/2}}{\sqrt{2}}\sum_{k=n}^{2n}\dfrac{1}{\sqrt{k}}. \end{align*} Now we see that \begin{align*} \sum_{k=n}^{2n}\dfrac{1}{\sqrt{k}}\approx\int_{n}^{2n}\dfrac{1}{\sqrt{x}}dx\approx\sqrt{n}\rightarrow\infty \end{align*} as $n\rightarrow\infty$.